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

The Second Derivative Test, and Connecting f, f′ and f″

B1.3 B1.2 B1.1About 70 minutes

Learning goals

I can use the second derivative test to classify critical points.
I can recognize when the second derivative test is inconclusive and use the first derivative test instead.
I can connect the key features of f with the features of the graphs of f′ and f″.

Words to know

second derivative test
If f′(c) = 0, then f″(c) > 0 means a local minimum and f″(c) < 0 means a local maximum. If f″(c) = 0 the test gives no conclusion.

inconclusive
Gives no answer either way; another method is needed.

A faster test for turning points

The first derivative test needs the sign of f′ on both sides of each critical number. If you already have f″, there is often a quicker way: check the concavity at the critical number. A flat spot on a smiling curve must be a valley. A flat spot on a frowning curve must be a hilltop.

The second derivative test

Suppose f′(c) = 0 and f″ exists near c.

If f″(c) > 0, f has a local minimum at c (concave up).
If f″(c) < 0, f has a local maximum at c (concave down).
If f″(c) = 0, the test is inconclusive: go back to the first derivative test.

The test does not apply where f′(c) does not exist (corners and cusps); use the first derivative test there too.

A horizontal tangent on a concave-up curve can only be at the bottom of a valley; on a concave-down curve it can only be at the top of a hill. That is the second derivative test.

Worked example

Example 1: Using the second derivative test

Find and classify the local extrema of f(x) = x³ − 12x + 1.

f′(x) = 3x² − 12 = 3(x − 2)(x + 2), so the critical numbers are x = ±2.
f″(x) = 6x.
f″(2) = 12 > 0, so a local minimum: f(2) = 8 − 24 + 1 = −15.
f″(−2) = −12 < 0, so a local maximum: f(−2) = −8 + 24 + 1 = 17.

Local maximum (−2, 17); local minimum (2, −15).

Worked example

Example 2: When the test is inconclusive

Classify the critical points of f(x) = x⁴ − 4x³.

f′(x) = 4x³ − 12x² = 4x²(x − 3): critical numbers 0 and 3.
f″(x) = 12x² − 24x. f″(3) = 108 − 72 = 36 > 0: local minimum, f(3) = −27.
f″(0) = 0: inconclusive. First derivative test: f′ = 4x²(x − 3) is negative on both sides of 0, so there is no extremum at 0.
In fact f″ = 12x(x − 2) changes sign at 0, so (0, 0) is a horizontal point of inflection.

Local minimum (3, −27). At (0, 0): no extremum (a horizontal inflection point).

Why f″(c) = 0 tells you nothing

Compare three functions, all with f′(0) = 0 and f″(0) = 0:

f(x) = x⁴ has a minimum at 0.
f(x) = −x⁴ has a maximum at 0.
f(x) = x³ has neither.

So when f″(c) = 0, every outcome is possible. Do not guess; use the first derivative test.

Worked example

Example 3: A rational function

Use the second derivative test to classify the critical points of f(x) = x + 1/x.

f′(x) = 1 − x⁻², which is 0 when x² = 1, so x = ±1.
f″(x) = 2x⁻³ = 2/x³.
f″(1) = 2 > 0: local minimum at (1, 2). f″(−1) = −2 < 0: local maximum at (−1, −2).

Local maximum (−1, −2); local minimum (1, 2).

Connecting f, f′ and f″

Here is the whole picture in one place. Read each row across.

On the graph of f	f′	f″
increasing	positive (above axis)	any sign
decreasing	negative (below axis)	any sign
local max or min (smooth)	zero, changing sign	usually negative at a max, positive at a min
concave up	increasing	positive
concave down	decreasing	negative
point of inflection	local max or min of f′	zero or undefined, changing sign
horizontal inflection point	zero, no sign change	changing sign

Each feature of f appears one level up on f′ and two levels up on f″.

Dots line up vertically. The turning points of f (x = 0, ±√3) are the zeros of f′. The inflection points of f (x = ±1) are the turning points of f′ and the zeros of f″.

Worked example

Example 4: Reading all three graphs

For f(x) = x⁴ − 6x², use the three graphs above to describe every key feature, then confirm with algebra.

f′ = 4x³ − 12x = 4x(x² − 3) has zeros at 0, ±√3 and signs −, +, −, +: minimums at x = ±√3, maximum at 0.
f″ = 12x² − 12: f″(±√3) = 36 − 12 = 24 > 0 (minimums ✓), f″(0) = −12 < 0 (maximum ✓).
f″ = 0 at x = ±1 with sign changes: inflection points. On the graph of f′ these are its turning points (−1, 8) and (1, −8).
Values: f(0) = 0, f(±√3) = 9 − 18 = −9, f(±1) = −5.

Local max (0, 0); local mins (±√3, −9); inflection points (±1, −5); concave up for |x| > 1, concave down for |x| < 1.

Just for fun

What did f″ say to a critical point at the bottom of a valley?

"Smile! You're a minimum."

Hands-on activity

Triple match-up

You need: 9 index cards or paper squares, a pencil, and a partner if possible.

Choose three functions: f(x) = x², g(x) = x³ − 3x and h(x) = −x⁴ + 2x². On three cards, sketch each function. On three more, sketch each first derivative (2x, 3x² − 3, −4x³ + 4x). On the last three, sketch each second derivative (2, 6x, −12x² + 4). Do not write labels on the cards.
Shuffle and hand the cards to your partner (or shuffle them yourself and wait a few minutes).
Sort the cards into three columns f, f′, f″ and match each row, explaining one reason for every match (for example, "this f′ crosses zero where that f turns").

What to look for: the horizontal line (f″ = 2) belongs to the parabola; the slanted line (6x) belongs to the cubic; the downward parabola (−12x² + 4) belongs to h. Every zero of an f′ card should line up with a turning point on its f card.

Your turn

Use the second derivative test to classify the critical points of f(x) = 2x³ − 3x² − 12x.

Classify the critical point of f(x) = x² − 8x + 3 with the second derivative test.

f′(5) = 0 and f″(5) = −3. What can you conclude?

f′(1) = 0 and f″(1) = 0. What can you conclude?

Classify the critical points of f(x) = x⁵ − 5x.

Classify the critical points of f(x) = 4/x + x.

More practice: warm-up

f′(c) = 0 and f″(c) > 0. Max or min?

f′(c) = 0 and f″(c) < 0. Max or min?

Name a function with f′(0) = 0, f″(0) = 0 and a minimum at 0.

On the graph of f′, what does a zero crossing from − to + mean for f?

If f″ > 0 on an interval, what is f′ doing there?

More practice: on your own

Classify all critical points of f(x) = 3x⁴ − 8x³ + 6x².

Classify the critical points of f(x) = x³ − 3x² − 9x + 5 and find the inflection point.

The graph of f′ is a line crossing the x-axis at x = 2 with slope −3. Classify x = 2.

f(x) = x³ − 6x² + 12x. Use both tests on its critical number.

Explain why the second derivative test cannot classify the critical point of f(x) = |x|.

Challenge

Find k so that f(x) = x³ + kx² + 3x has a critical number at x = 1, then classify it.

f(x) = ax⁴ + bx² (a ≠ 0, b ≠ 0) has a critical number at 0. Use the second derivative test to decide when it is a max and when a min.

Sketch, in words, a function with f(2) = 3, f′(2) = 0 and f″(x) < 0 for all x. How many local extrema can it have?

Check your understanding

Quiz

f′(3) = 0 and f″(3) = 7. What is at x = 3?

For f(x) = x³ − 27x, which point is a local maximum?

Which example shows that f″(c) = 0 can happen at a maximum?

A point of inflection of f appears on the graph of f″ as:

The graph of f′ is above the x-axis and rising on an interval. On that interval, f is:

For f(x) = x⁴ − 4x³, the second derivative test at x = 0 is:

For parents and teachers

Big idea (B1.3, B1.2, B1.1): at a horizontal tangent, the sign of f″ tells you whether you are at a valley (f″ > 0) or a hilltop (f″ < 0). When f″ = 0, the test says nothing, and your teen must fall back on the first derivative test. The summary table in this lesson connects every feature of f to f′ and f″.

How to help: cover one column of the summary table and ask your teen to fill it in from the others. The card match-up is a good review game.

Common difficulties:

Concluding "inflection point" whenever f″(c) = 0 at a critical number. It might be a max, a min or neither.
Applying the test where f′(c) does not exist.
Mixing up which graph a feature belongs to: a zero of f′ is a turning point of f, a zero of f″ (with sign change) is an inflection point of f.