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B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Local Maximums and Minimums: The First Derivative Test

B1.3 B1.5About 70 minutes
Learning goals
  • I can explain why local extrema occur only at critical numbers.
  • I can use the first derivative test to classify each critical point as a local maximum, local minimum or neither.
  • I can find absolute extrema of a continuous function on a closed interval.

Words to know

local maximum
A point that is at least as high as every point near it.
local minimum
A point that is at least as low as every point near it.
extremum (plural extrema)
A maximum or a minimum.
first derivative test
Classifies a critical number by how the sign of f′ changes across it: + to − is a maximum, − to + is a minimum, no change is neither.
absolute maximum / minimum
The greatest / least value of f on its whole domain or on a given interval.

Hilltops and valley floors

A snowboarder wants to know where the highest points of a run are. A business wants the production level with the biggest profit. Both are asking about extreme values. In the last lesson you found where a function rises and falls. Now you will use that to find and classify its local maximums and minimums.

**Remember from MHF4U (Polynomial Functions and Equations): a polynomial of degree n has at most n − 1 turning points. Calculus now tells you exactly** where they are.

Definitions

  • f has a local maximum at c if f(c) ≥ f(x) for all x near c. The value f(c) is a local maximum value.
  • f has a local minimum at c if f(c) ≤ f(x) for all x near c.
  • Together these are local extrema (singular: extremum).

Fermat's theorem: if f has a local extremum at c, then c is a critical number: either f′(c) = 0 or f′(c) does not exist. The reverse is false: a critical number is only a candidate. You need a test.

The first derivative test

Suppose c is a critical number and f is continuous at c.

  • If f′ changes from positive to negative at c, f has a local maximum at c.
  • If f′ changes from negative to positive at c, f has a local minimum at c.
  • If f′ does not change sign at c, f has no local extremum at c.

Think of walking along the curve: uphill then downhill means you passed a hilltop.

+ then −: local maxxy−11−11− then +: local minxy−11−11+ then +: neitherxy−11−11
The first derivative test looks only at the sign of f′ on each side of a critical number. A change from + to − gives a local maximum, − to + gives a local minimum, and no change gives neither.
Worked example

Example 1: Classifying turning points of a cubic

Find and classify the local extrema of f(x) = x³ − 6x² + 9x + 2.

  1. f′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).
  2. Critical numbers: x = 1 and x = 3.
  3. Signs: for x < 1, both factors negative, f′ > 0. For 1 < x < 3, f′ < 0. For x > 3, f′ > 0.
  4. At x = 1: + to −, so a local maximum. f(1) = 1 − 6 + 9 + 2 = 6.
  5. At x = 3: − to +, so a local minimum. f(3) = 27 − 54 + 27 + 2 = 2.

Local maximum (1, 6) and local minimum (3, 2).

Worked example

Example 2: A quartic with three turning points

Find and classify the local extrema of f(x) = 3x⁴ − 4x³ − 12x² + 5.

  1. f′(x) = 12x³ − 12x² − 24x = 12x(x² − x − 2) = 12x(x − 2)(x + 1).
  2. Critical numbers: x = −1, 0, 2.
  3. Build the sign chart below. The signs of f′ are −, +, −, + on the four intervals.
  4. f(−1) = 3 + 4 − 12 + 5 = 0. f(0) = 5. f(2) = 48 − 32 − 48 + 5 = −27.

Local minimums at (−1, 0) and (2, −27); local maximum at (0, 5).

f′(x) = 12x(x + 1)(x − 2)x−10212x−−++0x + 1−+++0x − 2−−−+0f′(x)−+−+000f(x)minmaxmin
The sign of f′ changes at all three critical numbers, so each one is a turning point: minimum, maximum, minimum.
y = 3x⁴ − 4x³ − 12x² + 5xy−112−30−20−105(−1, 0)(0, 5)(2, −27)
The quartic has a local minimum at (−1, 0), a local maximum at (0, 5) and a local minimum at (2, −27). The lowest point overall is (2, −27).
Worked example

Example 3: A critical number where f′ does not exist

Find and classify the critical points of f(x) = x^(2/3).

  1. f′(x) = (2/3)x^(−1/3) = 2/(3x^(1/3)).
  2. f′ is never 0, and it does not exist at x = 0. Since f(0) = 0 is defined, x = 0 is a critical number.
  3. For x < 0, x^(1/3) < 0 so f′ < 0. For x > 0, f′ > 0.
  4. − to +: local minimum at (0, 0). The graph has a sharp cusp there, with tangents becoming vertical.

Local minimum at (0, 0), at a cusp where f′ does not exist.

Absolute extrema on a closed interval

Often x is limited to an interval a ≤ x ≤ b (time cannot be negative; a fence cannot be longer than the material). On a closed interval, a continuous function always has an absolute maximum and an absolute minimum (the Extreme Value Theorem). They occur either at a critical number inside the interval or at an endpoint.

Closed interval method:

  1. Find the critical numbers of f inside (a, b).
  2. Evaluate f at each of them and at both endpoints a and b.
  3. The largest value is the absolute maximum; the smallest is the absolute minimum.

You will use this method constantly in the optimization lessons of Applications of Derivatives.

y = x^(2/3)xy−2−11212cusp (0, 0)y = x³ − 3x on [−2, 3]xy−2−1123−551015(−2, −2)(−1, 2)(1, −2)(3, 18)
Left: at a cusp f′ does not exist, yet the point is still a local minimum (f′ changes from − to +). Right: on a closed interval, the absolute maximum (3, 18) is at an endpoint, and the absolute minimum value −2 occurs twice.
Worked example

Example 4: Closed interval method

Find the absolute maximum and minimum of f(x) = x³ − 3x on −2 ≤ x ≤ 3.

  1. f′(x) = 3x² − 3 = 3(x − 1)(x + 1). Critical numbers −1 and 1, both inside the interval.
  2. Evaluate: f(−2) = −8 + 6 = −2; f(−1) = −1 + 3 = 2; f(1) = 1 − 3 = −2; f(3) = 27 − 9 = 18.

Absolute maximum 18 at x = 3 (an endpoint). Absolute minimum −2, reached at both x = −2 and x = 1. Note that the local maximum (−1, 2) is not the absolute maximum.

Just for fun

Riddle: Why did the local maximum refuse to brag?

Because it knew it was only the best in the neighbourhood.

Hands-on activity

Hill profile walk

You need: a measuring tape or a long string marked every 50 cm, a notebook, a pencil, and a nearby slope, snowbank or set of front steps (stay off icy surfaces and roads; have a grown-up come along).

  1. Lay the tape along the ground over a bump or slope and mark distance x every 50 cm.
  2. At each mark, describe the ground as up, down or flat as you face forward. This is your sign of f′.
  3. Sketch a height profile from your signs. Mark where the signs switch from up to down (local max) or down to up (local min).
  4. If your path has endpoints higher than any bump, note that the absolute maximum on your "interval" may be at an end.

What to look for: every local max on your sketch should sit between an "up" mark and a "down" mark, and every local min between "down" and "up". A flat stretch between two "up" marks is a critical spot that is not a turning point.

Your turn

1

Find and classify the local extrema of f(x) = x³ − 12x.

2

Find and classify the local extrema of f(x) = −2x² + 8x − 1.

3

Find and classify the local extrema of f(x) = x⁴ − 2x².

4

Does f(x) = x³ + 3x² + 3x have any local extrema?

5

f′(x) = (x + 2)(x − 1)². Classify each critical number.

6

Find the absolute extrema of f(x) = x² − 4x + 1 on 0 ≤ x ≤ 5.

More practice: warm-up

1

f′ changes from − to + at x = 4. What happens at x = 4?

2

f′ changes from + to − at x = −1. What happens at x = −1?

3

f′ is positive on both sides of x = 0. Is there an extremum at 0?

4

Find the critical number of f(x) = x² + 6x.

5

On a closed interval, name the two kinds of places where absolute extrema can occur.

More practice: on your own

1

Find and classify the local extrema of f(x) = 2x³ + 3x² − 36x.

2

Find and classify the local extrema of f(x) = x⁴ − 4x³ + 4x².

3

Find and classify the local extrema of f(x) = x + 1/x.

4

Find the absolute extrema of f(x) = x³ − 3x² on −1 ≤ x ≤ 4.

5

Classify the critical point of f(x) = (x − 3)^(2/3) + 1.

Challenge

1

Find a and b so that f(x) = x³ + ax² + b has a local minimum at (2, −1).

2

Prove that a cubic f(x) = ax³ + bx² + cx + d (a ≠ 0) has two local extrema if b² > 3ac and none if b² ≤ 3ac.

3

Find the absolute extrema of f(x) = x/(x² + 4) on 0 ≤ x ≤ 5.

Check your understanding

Quiz

1.

f′ changes from negative to positive at x = 3. What does f have at x = 3?

2.

Which function has a critical number at x = 0 that is NOT a local extremum?

3.

What is the local maximum point of f(x) = x³ − 3x?

4.

On 0 ≤ x ≤ 4, what is the absolute maximum of f(x) = x² − 2x?

5.

Fermat's theorem says that at a local extremum c:

6.

f′(x) = −(x − 5)³. What is at x = 5?

For parents and teachers

Big idea (B1.3, B1.5): turning points can only occur at critical numbers, and the first derivative test decides whether each one is a maximum, a minimum or neither. On a closed interval, also check the endpoints for absolute extrema.

How to help: have your teen say each step aloud: "find f′, factor, critical numbers, sign chart, classify, find the y-values." Forgetting the last step (computing f(c)) is very common.

Common difficulties:

  • Reporting only x = c instead of the point (c, f(c)).
  • Substituting into f′ instead of f to get the y-coordinate.
  • Forgetting endpoints in closed-interval problems.
  • Calling every critical number a turning point.