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The derivative becomes a toolkit

◆FRAME

Chapter 5 built the derivative $f'(x)$ as a limit — a single number attached to a curve at a point, describing how steep it is there. This chapter stops asking what a derivative IS and starts asking what it is FOR.

Two questions recur through it, and every one is answered by reading the same $f'(x)$ differently. Is a quantity growing or shrinking right now, and how fast? Where does a function peak, dip, or level off over an interval?

*Neither needs a new derivative — each one is the SAME $f'(x)$ from Chapter 5, put to a different practical use.* Rates, maxima and minima, optimisation: three names for three ways of reading one number.

Nothing in this chapter is a new object, which is worth knowing at the start. The three names ahead — rates of change, maxima and minima, optimisation — sound like three topics and are three uses of the one thing Chapter 5 already built.

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Rate of change via the chain rule

▲CONCEPT

When $y$ depends on $x$, and $x$ itself depends on $t$, a change in $t$ ripples through $x$ before it ever reaches $y$. The chain rule tracks that ripple: $(d y)/(d t) = (d y)/(d x) \cdot (d x)/(d t)$.

Read it plainly: the rate of change of $y$ with respect to time is the rate of change of $y$ with respect to $x$, scaled by how fast $x$ itself is changing. Neither factor alone answers the question — both are needed, multiplied together.

A purely geometric derivative turns physical the instant its variable starts moving with time. $(d V)/(d r)$, a sphere’s volume against its radius, is pure geometry — no clock involved. The moment the radius itself grows with time, the same derivative becomes the physical rate $(d V)/(d t) = (d V)/(d r) \cdot (d r)/(d t)$, and the chain rule is exactly the tool that makes that substitution legal.

Related rates are always this shape: one rate is given, another is wanted, and a geometric relation between the two quantities links them once it has been differentiated. Write the relation down first and the chain rule second — the arithmetic is never the hard part here.
Worked example

Find $(d r)/(d t)$ for a balloon’s radius when its volume grows at $900$ $\text{cm}^3$/s

  1. A spherical balloon’s volume increases at $900$ $\text{cm}^3$ per second; find $(d r)/(d t)$ when $r = 15$ cm
    Given rate; the chain rule connects the volume rate to the radius rate.
  2. $V = (4/3) \pi r^3$
    The volume of a sphere in terms of its radius.
  3. $(d V)/(d t) = 4 \pi r^2 \cdot (d r)/(d t)$
    Differentiate both sides with respect to $t$, applying the chain rule to $r^3$.
  4. $900 = 4 \pi (15)^2 \cdot (d r)/(d t) = 900 \pi \cdot (d r)/(d t)$
    Substitute the given rate $(d V)/(d t) = 900$ and $r = 15$.
  5. *$(d r)/(d t) = 1/\pi$ cm/s, about $0.32$ cm/s*
    Divide both sides by $900 \pi$ — the radius grows slower than it looks, since the same volume increase now stretches over a much larger surface.

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Reading the sign of the derivative

▲CONCEPT
The qualifier in the rule is doing real work, and this is where it bites. Where a derivative vanishes at a scattering of isolated points, the function is still climbing across the whole interval. So an answer that reports those points as places where it stops has read the rule too literally.
The method scales in a way that sketching does not. Two sign changes give three stretches, ten give eleven, and the work is the same each time. Solve for the zeros, test one point inside each gap, and read the shape off the signs.

A function’s derivative carries more than a number at each point — its SIGN tells a story about the whole neighbourhood around that point. $f$ is increasing on an interval where $f'(x) > 0$ throughout: the curve climbs as $x$ moves right. $f$ is decreasing where $f'(x) < 0$ throughout: the curve falls.

Where $f'(x) = 0$ across a whole sub-interval — not just at one isolated point — $f$ is constant there; nothing is climbing or falling at all.

This test reads off climbing versus falling from the sign of the derivative alone, without plotting a single point of the curve itself. Compute $f'(x)$ once, split the domain at the points where it changes sign, and the picture of the whole curve’s rise and fall follows from arithmetic, not a sketch.

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Local max/min — two tests

▲CONCEPT
The third outcome is the one that gets dropped. A candidate list is not an answer list, and a solution that reports every place the derivative vanishes as a turning point has skipped the only step that does any deciding.

A CRITICAL POINT of $f$ is a point $a$ where $f'(a) = 0$, or where $f'(a)$ fails to exist at all — a candidate location for a peak or a dip, nothing more yet.

The FIRST DERIVATIVE TEST turns that candidate into a verdict by watching how $f'$ changes SIGN on either side of $a$. A derivative that switches from positive to negative marks a local maximum: the function was climbing, and starts falling. A switch from negative to positive marks a local minimum: falling, then climbing.

A third outcome is just as real as the first two: no sign change at all. $f'$ stays positive on both sides, or negative on both sides, and $a$ is neither a max nor a min — the function was already climbing (or falling) straight through it.

▲CONCEPT
A test that can decline to answer is not a broken test; it is an honest one. What matters is knowing what to do when it declines — go back to the sign change on either side, which always decides. That is why the first test is never made obsolete by the second.

At a critical point $x = a$ where $f'(a) = 0$, a second route to the same verdict skips the sign-change walk entirely and checks a single number instead: $f''(a)$, the second derivative at that point.

*$f''(a) > 0$ means $a$ is a local minimum* — the curve bends upward there, concave up, cupping the point from below. $f''(a) < 0$ means a local maximum — concave down, the curve arching over the point from above.

One case escapes the test rather than answering it: $f''(a) = 0$. Concavity alone gives no verdict here — the SECOND DERIVATIVE TEST falls silent, and the first derivative test’s sign-change walk is the fallback that still works.

Two tests, one answer, and a choice to make under time pressure. The second derivative is quicker when it is easy to compute and does not come out zero; the walk across the sign change is slower but never declines to answer. Knowing which is cheaper for the function in front of you is the practical skill.
Worked example

Classify the critical points of $f(x) = x^3 - 6x^2 + 9x + 15$ using both tests

  1. $f(x) = x^3 - 6x^2 + 9x + 15$; find and classify its critical points
    Given cubic; find where $f'(x) = 0$ first.
  2. $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$
    Differentiate and factor; the critical points are where this equals zero.
  3. Critical points at $x = 1$ and $x = 3$
    The two roots of the factored derivative.
  4. First derivative test — $f'$ is positive before $x=1$, negative between $1$ and $3$, positive after $3$
    Test the sign of $f'(x) = 3(x-1)(x-3)$ in each of the three intervals the two roots split the line into.
  5. $x = 1$ is a local max (positive to negative); $x = 3$ is a local min (negative to positive)
    Read the verdict straight off the two sign changes.
  6. Second derivative test — $f''(x) = 6x - 12$
    Differentiate $f'(x)$ once more.
  7. $f''(1) = -6 < 0$ confirms a max; $f''(3) = 6 > 0$ confirms a min
    Evaluate the second derivative at each critical point independently.
  8. *Both tests agree: local max value $f(1) = 19$, local min value $f(3) = 15$*
    Substitute $x=1$ and $x=3$ back into $f$ itself to get the actual heights, not just their classification.
✕MISCONCEPTION
One counterexample is enough to kill a one-way inference, and this is the one to keep. Every turning point sits where the derivative vanishes; that does not make every such place a turning point, and no amount of further algebra will tell you which you have.

$f'(a) = 0$ is the FIRST DERIVATIVE TEST’s own starting condition for a critical point, which invites a shortcut: since every local max or min sits at a zero of $f'$, it can look as though every zero of $f'$ must sit at a max or min.

That reasoning skips a step the test never promised to skip. $f'(a) = 0$ only says the tangent is momentarily flat; it says nothing on its own about whether $f$ turns around there or simply pauses and keeps going the same way.

$f(x) = x^3$ at $x = 0$ makes the gap concrete. $f'(0) = 0$ — the tangent at the origin is flat. But $f'(x) = 3x^2 \geq 0$ EVERYWHERE, never negative, so $f$ never stops climbing; it only slows to a momentary halt at $x=0$ before continuing upward. $x = 0$ is an INFLECTION POINT, not a max or a min.

*A zero of $f'$ is only a candidate, never a verdict* — the sign change on either side, or the value of $f''$ there, still has to be tested before a critical point earns the name extremum.

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Absolute max/min on a closed interval

▲CONCEPT
The comparison list is the whole method, and it is short: every place the derivative vanishes, plus both ends of the interval. Evaluate the function at all of them and compare the numbers. There is no need to work out which are maxima and which are minima before answering.

Local tests answer a narrower question than they sound like they do: a local max is only the highest point compared to its immediate neighbours, not necessarily the highest point on the whole domain.

On a CLOSED interval $[a, b]$, a continuous $f$ is guaranteed to have both an absolute maximum and an absolute minimum somewhere on it — the Extreme Value Theorem promises this outright, no extra work needed to know one exists.

Finding them needs one combined list, not two separate searches: every critical point of $f$ strictly inside $(a, b)$, PLUS both endpoints $a$ and $b$. Evaluate $f$ at every entry on that list and compare. The largest value is the absolute maximum; the smallest is the absolute minimum — nowhere else on $[a, b]$ can either extreme hide.

✕MISCONCEPTION
This is the most reliable way to lose marks in the chapter, and it is not a calculus error — the calculus was right. On a closed interval the candidate list has to include both ends, and the derivative will never hand them to you, because they were never candidates by its standard.

Absolute extrema share their procedure’s first half with local extrema — find the critical points, evaluate $f$ there — which invites treating the two searches as identical. If critical points already located every local peak and dip, it can look like nothing more is needed.

That shortcut drops half the required list. A closed interval’s boundary is not scenery — an endpoint is a legitimate candidate for the absolute maximum or minimum, even though it is not, and never can be, a critical point at all.

$f(x) = x^2$ on $[-1, 2]$ makes the gap concrete. Its only critical point is $x = 0$, which gives the local AND absolute minimum, value $0$. But the absolute MAXIMUM sits at the endpoint $x = 2$, value $4$ — a point the critical-point search never visits, because a function is under no obligation to have zero derivative at the edge of its domain.

Skipping the endpoints does not just risk missing a tie — it can miss the actual maximum outright.

Notice that the two local extremes were not the answer and were still worth finding — they had to be on the list before they could be ruled off it. That is the whole difference between a local question and an absolute one, and it is what the two extra evaluations buy.
Worked example

Find the absolute max and min of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$

  1. $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$; find the absolute max and min
    Given cubic on a closed interval; list critical points and both endpoints.
  2. $f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$
    Differentiate and factor to locate the critical points.
  3. Critical points at $x = 2$ and $x = 3$, both inside $(1, 5)$
    The two roots of the factored derivative; both fall within the open interval.
  4. $f(1) = 24$, $f(2) = 29$, $f(3) = 28$, $f(5) = 56$
    Evaluate $f$ at every critical point AND both endpoints — the full comparison list.
  5. *Absolute maximum $56$ at $x=5$; absolute minimum $24$ at $x=1$ — both at the endpoints*
    The largest and smallest values on the list decide it; the interior critical values ($29$, $28$) turn out to be only LOCAL extrema, not absolute ones.

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Optimisation

▲CONCEPT

A practical optimisation problem never hands over a bare function to maximize or minimize — it hands over a quantity, tied to some CONSTRAINT that links several variables together.

The method turns a many-variable problem into a one-variable one before calculus ever gets involved: use the constraint to write every other variable in terms of a single one, so the quantity to optimise becomes a function of that one variable alone.

From there the earlier toolkit takes over unchanged. Differentiate, set the derivative to zero to find the critical points, and classify each one with the first or second derivative test — adding a check against the boundary as well, whenever the variable’s own domain is closed.

The hard part of an optimisation question is over before any calculus starts: it is spending the constraint to get down to a single variable. Do that badly and no amount of differentiating will rescue it; do it well and what is left is the routine from earlier in this chapter.
Worked example

Minimize a closed cylinder’s surface area for a fixed volume of $16 \pi$ $\text{cm}^3$

  1. A closed cylindrical can must hold volume $16 \pi$ $\text{cm}^3$; minimize its total surface area
    Given constraint (fixed volume) and quantity to minimize (surface area); express area in one variable.
  2. $\pi r^2 h = 16 \pi$, so $h = 16/r^2$
    The volume constraint, solved for $h$ in terms of $r$.
  3. $S(r) = 2 \pi r^2 + 2 \pi r h = 2 \pi r^2 + 32 \pi/r$
    Substitute $h$ into the surface-area formula, eliminating $h$ so $S$ depends only on $r$.
  4. $S'(r) = 4 \pi r - 32 \pi/r^2$
    Differentiate with respect to $r$ to find the critical point.
  5. Setting $S'(r) = 0$ gives $r^3 = 8$, so $r = 2$, and $h = 16/4 = 4$
    Solve the critical-point equation, then recover $h$ from the constraint.
  6. $S''(r) = 4 \pi + 64 \pi/r^3 > 0$ for every $r > 0$
    The second derivative test confirms this critical point is a minimum, not a maximum.
  7. *Minimum surface area $S = 24 \pi$ $\text{cm}^2$ at $r=2$, $h=4$*
    Substitute the optimal radius back into $S(r)$ — a two-variable can has shrunk to one number.

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One derivative, many readings

■RECAP
Worth coming back to when a question in the exercise set looks unfamiliar. Nearly every one of them asks for one of these readings, and working out which one is being asked for is usually more than half the work.

Every tool in this chapter reads the SAME derivative differently, never a new one. Its VALUE gives a rate ($(d y)/(d t)$). Its SIGN gives increasing or decreasing. Its ZERO-CROSSINGS flag CANDIDATE turning points — never proof.

*A zero of $f'$ is only ever a candidate, and it has to be tested, never trusted on sight* — the sign change on either side, or the value of $f''$ there, is what turns a candidate into a verdict. On a closed interval, the endpoints stay in the comparison right until the very last step; dropping them can drop the actual answer along with them.

Rates, maxima and minima, optimisation: three different practical questions, one derivative, read three different ways.

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Practice set

The same three-factor pattern with a different geometry, which is why it sits next to the balloon. Once it has been seen twice the exercise set stops being a list of different problems and becomes one problem with a different formula dropped into it.

Exercise 6.1 practises finding rates of change using the chain rule. Problems cover circles, spheres, and cubes — given how one quantity changes with time, find how a related quantity changes at the same instant.

Exercise 6.1 — Rate of change
  1. practice The radius of a circle is increasing at the rate of $4$ cm/s. Find the rate at which the area of the circle is increasing when the radius is $6$ cm.
  2. practice An edge of a variable cube is increasing at the rate of $2$ cm/s. How fast is the volume of the cube increasing when the edge is $9$ cm long?
  3. practice The radius of a circle is increasing at the rate of $0.5$ cm/s. What is the rate of increase of its circumference?
  4. practice The radius of a spherical soap bubble is increasing at the rate of $2$ cm/s. Find the rate at which its surface area is increasing when the radius is $6$ cm.
  5. practice The side of a variable square is increasing at the rate of $6$ cm/s. Find the rate of increase of its area when the side is $10$ cm.
  6. practice A spherical balloon has a variable radius. Find the rate of change of its volume with respect to the radius when the radius is $7$ cm.
  7. practice The length $x$ of a rectangle is decreasing at the rate of $4$ cm/minute and the width $y$ is increasing at the rate of $3$ cm/minute. When $x = 12$ cm and $y = 8$ cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.
  8. practice The radius of an air bubble is increasing at the rate of $1/3$ cm/s. At what rate is the volume of the bubble increasing when the radius is $2$ cm?
  9. practice A spherical balloon is being inflated by pumping in $850$ cubic centimetres of gas per second. Find the rate at which its radius is increasing when the radius is $17$ cm.
  10. practice Sand is pouring from a pipe at the rate of $20$ cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-fourth of the radius of the base. How fast is the height of the sand cone increasing when the height is $5$ cm?
  11. practice A ladder $13$ m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $3$ m/s. How fast is its height on the wall decreasing when the foot of the ladder is $5$ m away from the wall?
  12. practice A spherical balloon has a variable diameter $3 (2 x + 1)$. Find the rate of change of its volume with respect to $x$, when $x = 2$.
  13. practice Which of these is the rate of change of $x^3 - 2 x$ with respect to $x$ at $x = 3$?
    1. $23$
    2. $25$
    3. $27$
    4. $29$
  14. practice The rate of change of the area of a circle with respect to its radius $r$ at $r = 4$ cm is
    1. $6 \pi$
    2. $8 \pi$
    3. $10 \pi$
    4. $4 \pi$
  15. practice A cube’s edge is increasing at the rate of $2$ cm/s. The rate of increase of its surface area, when the edge is $5$ cm, is
    1. $100$
    2. $110$
    3. $120$
    4. $130$
  16. practice Two quantities $x$ and $y$ satisfy $x^2 - y^2 = 16$, and both change with time. If $(d x)/(d t) = 3$ units/s, find $(d y)/(d t)$ at the instant when $x = 5$ and $y = 3$.
  17. practice A particle moves along the curve $4 y = x^3 + 8$. Find the points on the curve at which the $y$-coordinate changes $12$ times as fast as the $x$-coordinate.
  18. practice A cube’s volume is increasing at the rate of $64$ cm³/s. Find the rate at which its surface area is increasing when the edge is $8$ cm.
  19. practice A kite is flying at a constant height of $90$ m above the ground and moving horizontally away from the person holding the string at the rate of $6$ m/s. Find the rate at which the string is being paid out when $150$ m of string is already out.
Answers
  1. $48 \pi$ cm²/s
  2. $486$ cm³/s
  3. $\pi$ cm/s
  4. $96 \pi$ cm²/s
  5. $120$ cm²/s
  6. $196 \pi$ cm²
  7. (a) perimeter decreasing at $2$ cm/min; (b) area increasing at $4$ cm²/min
  8. $(16 \pi)/3$ cm³/s
  9. $25/(34 \pi)$ cm/s
  10. $1/(20 \pi)$ cm/s
  11. $(d y)/(d t) = -5/4$ m/s — the height is decreasing.
  12. $675 \pi$
  13. B — $25$.
  14. B — $8 \pi$.
  15. C — $120$.
  16. $5$ units/s
  17. $(4, 18)$ and $(-4, -14)$
  18. $32$ cm²/s
  19. $4.8$ m/s

Exercise 6.2 practises finding where a given function increases or decreases. Find $f'(x)$, then read its sign on each interval the critical points split the domain into.

Exercise 6.2 — Increasing and decreasing functions
  1. practice Show that the function given by $f (x) = 4 x - 5$ is increasing on $\mathbb{R}$.
  2. practice Show that the function given by $f (x) = e^{3 x}$ is increasing on $\mathbb{R}$.
  3. practice Show that the function given by $f (x) = \log x$ is increasing on $(0, \infty)$.
  4. practice Find the intervals in which $f (x) = 3 x^2 - 12 x + 7$ is increasing or decreasing.
  5. practice Find the intervals in which $f (x) = 4 x^2 + 5 x - 3$ is increasing or decreasing.
  6. practice Find the intervals in which $f (x) = x^2 - 8 x + 12$ is increasing or decreasing.
  7. practice Show that $f (x) = \sin x$ is (a) increasing on $(0, \pi/2)$ and (b) decreasing on $(\pi/2, \pi)$.
  8. practice Find the intervals in which each of the following functions is strictly increasing or strictly decreasing: (a) $x^2 + 4 x - 5$ (b) $10 - 7 x - 2 x^2$.
  9. practice Prove that the function $f (x) = x^2 + x - 1$ is neither increasing nor decreasing on $(-1, 1)$.
  10. practice Which of the following functions is decreasing on $(0, \pi/2)$?
    1. $\sin x$
    2. $\tan x$
    3. $\cos x$
    4. $\sin 2 x$
  11. practice On which of the following intervals is the function $f (x) = x^60 + \sin x - 1$ decreasing?
    1. $(0, 1)$
    2. $(\pi/2, \pi)$
    3. $(0, \pi/2)$
    4. None of these
  12. practice Find the intervals in which $f (x) = 2 x^3 - 9 x^2 + 12 x + 5$ is increasing or decreasing.
  13. practice Find the intervals in which $f (x) = 2 x^3 - 3 x^2 - 36 x + 10$ is strictly increasing or strictly decreasing.
  14. practice Find the intervals in which $f (x) = -2 x^3 - 3 x^2 + 12 x + 5$ is strictly increasing or strictly decreasing.
  15. practice For what values of $a$ is the function $f (x) = x^2 + a x + 3$ increasing on the interval $[2, 4]$?
  16. practice Find the values of $x$ for which $f (x) = 2 x^3 - 15 x^2 + 24 x + 7$ is strictly decreasing.
  17. practice Find the values of $x$ for which $y = [x (x - 3)]^2$ is an increasing function.
  18. practice Find the intervals in which $f (x) = 3 x^4 - 4 x^3 - 12 x^2 + 5$ is increasing or decreasing.
  19. practice Find the intervals in which $f (x) = \sin 4 x$, $x \in (0, \pi/2)$, is increasing and decreasing.
  20. practice Find the values of $x$ for which $f (x) = (x + 1)^3 (x - 2)^3$ is an increasing function.
Answers
  1. Proved — $f'(x) = 4 > 0$ for every $x \in \mathbb{R}$, so $f$ is strictly increasing on $\mathbb{R}$.
  2. Proved — $f'(x) = 3 e^{3 x} > 0$ for every $x \in \mathbb{R}$, so $f$ is increasing on $\mathbb{R}$.
  3. Proved — $f'(x) = 1/x > 0$ for every $x > 0$, so $f$ is increasing on $(0, \infty)$.
  4. Decreasing on $(-\infty, 2)$; increasing on $(2, \infty)$.
  5. Decreasing on $(-\infty, -5/8)$; increasing on $(-5/8, \infty)$.
  6. Decreasing on $(-\infty, 4)$; increasing on $(4, \infty)$.
  7. Proved — $f'(x) = \cos x$ is positive on $(0, \pi/2)$ and negative on $(\pi/2, \pi)$.
  8. (a) decreasing on $(-\infty, -2)$, increasing on $(-2, \infty)$; (b) increasing on $(-\infty, -7/4)$, decreasing on $(-7/4, \infty)$.
  9. Proved — $f'(x) = 2 x + 1$ is negative on $(-1, -1/2)$ and positive on $(-1/2, 1)$, so $f$ is neither increasing nor decreasing throughout $(-1, 1)$.
  10. C — $\cos x$.
  11. D — None of these.
  12. Increasing on $(-\infty, 1) \cup (2, \infty)$; decreasing on $(1, 2)$.
  13. Increasing on $(-\infty, -2) \cup (3, \infty)$; decreasing on $(-2, 3)$.
  14. Decreasing on $(-\infty, -2) \cup (1, \infty)$; increasing on $(-2, 1)$.
  15. $a \geq -4$.
  16. $1 < x < 4$.
  17. Increasing on $(0, 3/2) \cup (3, \infty)$.
  18. Decreasing on $(-\infty, -1) \cup (0, 2)$; increasing on $(-1, 0) \cup (2, \infty)$.
  19. Increasing on $(0, \pi/8) \cup (3 \pi/8, \pi/2)$; decreasing on $(\pi/8, 3 \pi/8)$.
  20. Increasing on $x > 1/2$; decreasing on $x < 1/2$.

Exercise 6.3 practises finding local maxima and minima using the first and second derivative tests. It also covers absolute extrema on closed intervals and practical optimisation problems, so identifying which tool the problem calls for is itself part of the exercise.

Exercise 6.3 — Maxima and minima
  1. practice Find the maximum or minimum value, if any, of $f (x) = (3 x - 2)^2 + 5$.
  2. practice Find the maximum or minimum value, if any, of $g (x) = -(x + 3)^2 + 8$.
  3. practice Find the maximum or minimum value, if any, of $h (x) = 4 x^2 + 4 x + 7$.
  4. practice Find the maximum or minimum value, if any, of $f (x) = |x - 3| + 2$.
  5. practice Find the local maxima and local minima, if any, of $f (x) = x^3 - 3 x + 5$, and find the corresponding values.
  6. practice Find the local maxima and local minima of $g (x) = x^3 - 6 x^2 + 9 x + 4$, and find the corresponding values.
  7. practice Find the local maxima and local minima of $f (x) = 2 x^3 - 3 x^2 - 12 x + 8$, and find the corresponding values.
  8. practice Find the local maximum and local minimum values, if any, of $f (x) = x^4 - 4 x^3$.
  9. practice Determine whether $x = 1$ is a point of local maximum, local minimum, or neither, for $f (x) = x^3 - 3 x^2 + 3 x + 7$.
  10. practice Prove that $f (x) = e^x$ has neither a local maximum nor a local minimum.
  11. practice Prove that $g (x) = \log x$ has neither a local maximum nor a local minimum.
  12. practice Find the absolute maximum and minimum values of $f (x) = x^3$ on $[-3, 3]$.
  13. practice Find the absolute maximum and minimum values of $f (x) = 6 x - x^2$ on $[-1, 4]$.
  14. practice Find the absolute maximum and minimum values of $f (x) = (x - 2)^2 + 5$ on $[-1, 4]$.
  15. practice Find the maximum profit that a company can make, if the profit function is given by $p (x) = 42 + 60 x - 15 x^2$.
  16. practice Find the absolute maximum and minimum values of $f (x) = x^3 - 6 x^2 + 9 x + 2$ on $[0, 5]$.
  17. practice At what points in $[0, 2 \pi]$ does the function $\sin 2 x$ attain its maximum value?
  18. practice What is the maximum value of $f (x) = \sin x + \cos x$?
  19. practice Find the maximum value of $2 x^3 - 24 x + 50$ on $[1, 3]$. Find the maximum value of the same function on $[-3, -1]$.
  20. practice It is given that at $x = 2$, the function $f (x) = x^4 - 70 x^2 + a x + 9$ attains its maximum value on the interval $[0, 3]$. Find the value of $a$.
  21. practice Find the absolute maximum and minimum values of $f (x) = x + \sin 2 x$ on $[0, \pi]$.
  22. practice Find two positive numbers whose sum is $20$ and whose product is maximum.
  23. practice Find two positive numbers $x$ and $y$ such that $x + y = 16$ and $x y^3$ is maximum.
  24. practice Find two positive numbers whose sum is $18$ and the sum of whose cubes is minimum.
  25. practice A square piece of tin of side $24$ cm is to be made into a box without a top, by cutting a square of side $x$ cm from each corner and folding up the flaps. What value of $x$ makes the volume of the box maximum, and what is that maximum volume?
  26. practice A rectangular sheet of tin $16$ cm by $10$ cm is to be made into a box without a top, by cutting off a square of side $x$ cm from each corner and folding up the flaps. What value of $x$ makes the volume of the box maximum?
  27. practice A closed cylindrical can is to have a total surface area of $96 \pi$ cm². Find the radius and height that maximize its volume.
  28. practice A wire of length $20$ m is cut into two pieces. One piece is bent into a square and the other into a circle. What should be the length of each piece so that the combined area is minimum?
  29. practice The point on the curve $x^2 = 4 y$ which is nearest to the point $(0, 5)$ is
    1. $(2 \sqrt{3}, 3)$
    2. $(0, 0)$
    3. $(2, 1)$
    4. $(4, 4)$
Answers
  1. Minimum value $5$ at $x = 2/3$; no maximum value.
  2. Maximum value $8$ at $x = -3$; no minimum value.
  3. Minimum value $6$ at $x = -1/2$; no maximum value.
  4. Minimum value $2$ at $x = 3$; no maximum value.
  5. Local maximum $7$ at $x = -1$; local minimum $3$ at $x = 1$.
  6. Local maximum $8$ at $x = 1$; local minimum $4$ at $x = 3$.
  7. Local maximum $15$ at $x = -1$; local minimum $-12$ at $x = 2$.
  8. Local minimum $-27$ at $x = 3$; $x = 0$ is a critical point but not a local extremum.
  9. Neither — $x = 1$ is a point of inflection, since $f'(x) = 3 (x - 1)^2$ never changes sign.
  10. Proved — $f'(x) = e^x > 0$ for every $x$, so $f$ has no critical point and hence no local extremum.
  11. Proved — $g'(x) = 1/x \neq 0$ anywhere in the domain, so $g$ has no critical point.
  12. Absolute maximum $27$ at $x = 3$; absolute minimum $-27$ at $x = -3$.
  13. Absolute maximum $9$ at $x = 3$; absolute minimum $-7$ at $x = -1$.
  14. Absolute maximum $14$ at $x = -1$; absolute minimum $5$ at $x = 2$.
  15. Maximum profit ₹$102$, at $x = 2$.
  16. Absolute maximum $22$ at $x = 5$; absolute minimum $2$ (attained at both $x = 0$ and $x = 3$).
  17. Maximum value $1$, attained at $x = \pi/4$ and $x = 5 \pi/4$.
  18. $\sqrt{2}$.
  19. Maximum on $[1,3]$ is $32$ (at $x=3$); maximum on $[-3,-1]$ is $82$ (at $x=-2$).
  20. $a = 248$.
  21. Absolute maximum $\pi$ at $x = \pi$; absolute minimum $0$ at $x = 0$ — both at the endpoints.
  22. The numbers are $10$ and $10$; maximum product $100$.
  23. $x = 4$, $y = 12$; maximum value of $x y^3$ is $6912$.
  24. $x = 9$, $y = 9$; minimum sum of cubes $1458$.
  25. $x = 4$ cm; maximum volume $1024$ cm³.
  26. $x = 2$ cm; maximum volume $144$ cm³.
  27. $r = 4$ cm, $h = 8$ cm; maximum volume $128 \pi$ cm³.
  28. Square piece $= 80/(\pi + 4)$ m; circle piece $= 20 \pi/(\pi + 4)$ m.
  29. A — $(2 \sqrt{3}, 3)$.

The Miscellaneous Exercise mixes every tool from this chapter into single multi-step problems. A rate-of-change setup might end in an optimisation, or a monotonicity check might need the second-derivative test to close it out — no problem announces which technique applies.

Miscellaneous Exercise — Chapter 6
  1. practice Find the maximum value of $f (x) = (\log x)/x$.
  2. practice The two equal sides of an isosceles triangle with fixed base $10$ cm are decreasing at the rate of $2$ cm per second. How fast is the area decreasing when the two equal sides are also $10$ cm each?
  3. practice Find the intervals in which $f (x) = x^3 - 3 x^2 - 9 x + 5$ is increasing or decreasing.
  4. practice Find the intervals in which $f (x) = x/(x^2 + 1)$ is increasing or decreasing.
  5. practice Find the absolute maximum and minimum values of $f (x) = \sin^2 x - \cos x$ on $[0, \pi]$.
  6. practice Find, in terms of $R$, the height of the right circular cylinder of greatest volume that can be inscribed in a sphere of radius $R$, and find that maximum volume.
  7. practice An open rectangular tank with a square base is to have a volume of $64$ m³. The base costs ₹$40$ per m² and the sides cost ₹$20$ per m². Find the dimensions for which the cost is minimum, and find that minimum cost.
  8. practice A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $12$ m. Find the width and height of the rectangle so that the window admits the maximum possible light.
  9. practice The sum of the perimeters of a circle and a square is $k$, a constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.
  10. practice Prove that $f (x) = x^3 + x^2 + x + 1$ has neither a local maximum nor a local minimum.
  11. practice Find the point on the curve $y = x^2 + 3$ nearest to the point $(0, 6)$.
  12. practice A particle’s displacement, in metres, at time $t$ seconds is given by $s (t) = t^3 - 6 t^2 + 9 t$, $t \geq 0$. Find the interval(s) of time during which the particle moves in the positive direction, and find its velocity at the instant its acceleration is zero.
  13. practice Find the local maximum value, if any, of $f (x) = x \sqrt{1 - x}$, $x < 1$.
  14. practice A cylindrical tank of radius $7$ m is being filled with wheat at the rate of $154$ cubic metres per hour. The depth of the wheat is then increasing at the rate of
    1. $1$ m/h
    2. $0.5$ m/h
    3. $2$ m/h
    4. $1.5$ m/h
  15. practice Find the absolute maximum and minimum values of $f (x) = x^3 - 3 x$ on $[-2, 3]$, and state whether $x = 1$ is a point of local maximum or local minimum.
  16. practice For a cone of fixed slant height $l$, show that the semi-vertical angle at which the volume is maximum satisfies $\tan \theta = \sqrt{2}$.
Answers
  1. Maximum value $1/e$ at $x = e$.
  2. $(d A)/(d t) = -(20 \sqrt{3})/3$ cm²/s — the area is decreasing.
  3. Increasing on $(-\infty, -1) \cup (3, \infty)$; decreasing on $(-1, 3)$.
  4. Increasing on $(-1, 1)$; decreasing on $(-\infty, -1) \cup (1, \infty)$.
  5. Absolute maximum $5/4$ at $x = 2 \pi/3$; absolute minimum $-1$ at $x = 0$.
  6. $h = (2 R)/\sqrt{3}$; maximum volume $(4 \sqrt{3} \pi R^3)/9$.
  7. Base side $4$ m, height $4$ m; minimum cost ₹$1920$.
  8. Width $= 24/(4 + \pi)$ m, height $= 12/(4 + \pi)$ m.
  9. Proved — at the critical point $s = 2 r$ (with $r = k/(2 (4 + \pi))$, $s = k/(4 + \pi) = 2 r$), and the second derivative there is positive, confirming a minimum.
  10. Proved — $f'(x) = 3 x^2 + 2 x + 1$ has negative discriminant and positive leading coefficient, so $f'(x) > 0$ for every $x$; $f$ has no critical point at all.
  11. $(\sqrt{10}/2, 11/2)$ and $(-\sqrt{10}/2, 11/2)$.
  12. Positive direction for $0 \leq t < 1$ and $t > 3$; velocity when acceleration is zero ($t = 2$) is $-3$ m/s.
  13. Local maximum $(2 \sqrt{3})/9$ at $x = 2/3$; no local minimum.
  14. A — $1$ m/h.
  15. Absolute maximum $18$ at $x = 3$; absolute minimum $-2$ (at $x = -2$ and $x = 1$); $x = 1$ is a point of local minimum.
  16. Proved — with $r = l \sin \theta$ and $h = l \cos \theta$, $V (\theta) = (\pi l^3/3) \sin^2 \theta \cos \theta$; setting $V'(\theta) = 0$ gives $2 \cos^2 \theta = \sin^2 \theta$, i.e. $\tan^2 \theta = 2$, so $\tan \theta = \sqrt{2}$ at the maximum.

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