Maxima & Minima

This placeholder page contains a summary of how to do max/min problems in Calculus, and free problems for you to practice, each with a complete solution immediately available. Let's jump right in!

PROBLEM SOLVING STRATEGY: Maxima & Minima

This is not a recipe for you simply to follow, but rather a series of steps to help guide your thinking that will often work well.

Compute the function's derivative, $𝑓' (𝑥)$ .
Find all critical numbers c such that either:
- the derivative is zero: $𝑓' (𝑐) = 0$ ;
- or the derivative $𝑓' (𝑐)$ does not exist.
Make a list of the values c that you find.
(If you are examining a closed interval ( $a \le x \le b$ ), then you may be able to skip to step 4.)
For each critical point c, check whether f is a local maximum or minimum or neither. Use either the First Derivative Test or the Second Derivative Test.
A. First Derivative Test:
Check a convenient value that's less than c to see whether $𝑓'$ is positive or negative there; then check a convenient value that's a greater than c to see if $𝑓'$ is positive or negative there. As the figure below illustrates:
- if the derivative switches from positive to negative, you've found a local maximum.
- if it switches from negative to positive, you've found a local minimum.
- if it doesn't switch, it is neither a maximum nor a minimum; it is simply a point with a horizontal tangent (if $𝑓' (𝑐) = 0$ ), or vertical tangent (if $𝑓' (𝑐)$ is undefined).
B. Second Derivative Test:
Compute the function's second derivative, $𝑓 ″ (𝑥) .$ Then determine the sign of the second derivative at c, $𝑓 ″ (𝑐)$ . As the figure below illustrates:
- if $𝑓 ″ (𝑐)$ is negative (so the function is concave down there), then $𝑐$ is a local maximum.
- if it's positive (so the function is concave up there), then $𝑐$ is a local minimum.
If requested, compute the value(s) of $𝑓 (𝑐)$ to find the $𝑦$ -values of the maxima and/or minima.
If the function has endpoints ( $𝑎 \leq 𝑥 \leq 𝑏$ ), compute the values of $𝑓 (𝑎)$ and $𝑓 (𝑏)$ to see how they compare to the values of $𝑓 (𝑐)$ you found step 4. The global (or absolute) maximum or minimum may lie at one of these endpoints.

Of course the most important thing is to practice. Problem 1 steps you through the process to find the relative maximum and minimum of a function. Problem 2 then examines the same function, but now looks for the global maximum and minimum on an interval.

Practice Problem 1: Relative maximum and minimum

𝑓 (𝑥) = 2 𝑥 3 - 3 𝑥 2 - 12 𝑥 + 1

Consider the function $𝑓 (𝑥) = 2 𝑥 3 - 3 𝑥 2 - 12 𝑥 + 1$ .

(a)

Find the critical points of

𝑓

(b)

Use the First Derivative Test to determine whether each critical point is a local maximum, minimum, or neither.

(c)

Use the Second Derivative Test to verify your answer to part (b).

(a) $- 1, 2$
(b) $𝑥 = - 1$ : local maximum; $𝑥 = 2$ : local minimum
(c) See detailed solution.

Most people using the First Derivative Test find it helpful to create a small diagram like the one below to organize their thinking and results. We place the values of

𝑥

underneath the number line, marking the critical numbers we found earlier. Above the line we show information about

𝑓' (𝑥)

, starting with the fact that at these critical points

𝑓'

is zero.

$As described in text$

The figure nicely illustrates that we need to consider the sign of

𝑓'

in three regions: (I) to the left of

- 1

; (II) between

- 1

and 2; and (III) to the right of 2.

(I) An easy value to try to the left of -1 is

𝑥 = - 2

𝑓' (- 2) = 6 (- 4) (- 1) > 0

We actually don't care what the value of

𝑓' (- 2)

is; we only care whether it's positive or negative. Since it's positive, we put a "+" above that region on our line:

$As described in text$

We thus know that the function has a positive derivative, and so is increasing, on the interval

(- \infty, - 1)

.

(II) A convenient value between -1 and 2 is zero:

𝑓' (0) = 6 (- 2) (1) < 0

Again, we don't care about the value specifically; without any further work we see that it's negative, and so we put a "-" above that region.

< $As described in text$

We thus know that the function has a negative derivative, and so is decreasing, on the interval

(- 1, 2)

.

Furthermore, we now know that the critical point

𝑥 = - 1

is a relative maximum, according to the First Derivative Test, because

𝑓'

switches from positive to negative there.

$When f' switches from positive to negative, you have a relative maximum.$

(III) A convenient value to try to the right of 2 is

𝑥 = 3

𝑓' (3) = 6 (1) (4) > 0

This value is positive, and so we put a "+" over that region on the line.

$As described in text.$

We now know that, after

𝑥 = 2

, the function always increases: it is increasing on

(2, \infty)

.

Furthermore, we conclude from the First Derivative Test that

𝑥 = 2

is a relative minimum, since

𝑓'

switches from negative to positive there.

$When f' switches from negative to positive, you have a relative minimum.$

Summarizing our results, we conclude our First Derivative Tests:

∙

Since

𝑓'

switches from positive to negative at

𝑥 = - 1

, this point is a local maximum.

✓

∙

Since

𝑓'

switches from negative to positive at

𝑥 = 2

, this point is a local minimum.

✓

For comparison, the figure below shows the graph of the function.

$As described in text$

Practice Problem 2: ... now absolute max and min on a closed interval

Consider again the function from Question 1,

𝑓 (𝑥) = 2 𝑥 3 - 3 𝑥 2 - 12 𝑥 + 1

.
Find the

𝑥

-values of absolute maximum and the absolute minimum on the interval [0, 3].

Practice Problem 3:

𝑓 (𝑥) = 𝑒 - 𝑥 s i n (𝑥)

Consider the function $𝑓 (𝑥) = 𝑒 - 𝑥 s i n 𝑥$ on the interval $[0, 2 𝜋]$ .

(a)

Find the critical points of

𝑓

(b)

Find the points of global maximum and minimum.

Practice Problem 4:

𝑓 (𝑥) = 𝑥 - l n (𝑥)

Consider the function

𝑓 (𝑥) = 𝑥 - l n 𝑥

on the interval [0.1, 2]. Find the values of

𝑥

for which

𝑓

has its global maximum and minimum.

Practice Problem 5: Find

𝑘

such that ... (Based on an actual exam question)

Find the value of

𝑘

such that

𝑓 (𝑥) = 𝑥 + 𝑘 𝑥

has a local minimum at

𝑥 = 3

Practice Problem 6: Find

𝑎

𝑏

such that ... (Based on an actual exam question)

Find values

𝑎

and

𝑏

so that the function

𝑓 (𝑥) = 𝑎 𝑥 𝑒 - 𝑏 𝑥

has a local maximum at the point (2, 4).

Practice Problem 7: A particle's motion

A particle moves along the x-axis so that its velocity at time $𝑡$ is given by $𝑣 (𝑡) = 𝑡 2 - 4 𝑡 + 3$ .

(a)

When is the particle at rest?

(b)

When is the particle moving to the right?

(c)

When does the particle change direction?

(d)

What is the farthest to the left of the origin that the particle moves?

(a) $𝑡 = 2, 5$
(b) $𝑡 > 5$
(c) $𝑡 = 5$
(d) -27

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