The second derivative test and extreme values on an interval

What you will be able to do

• use the second derivative test, and say when it fails;
• explain what the sign of $f''$ says about the shape of the graph;
• find the greatest and least values of a function on a closed interval;
• find the extreme values of a function on an open or an unbounded interval.

What the second derivative says

The second derivative $f''$ is the rate of change of the gradient $f'$.

• If $f'' > 0$, the gradient is increasing. The graph bends upwards, like a cup.
• If $f'' < 0$, the gradient is decreasing. The graph bends downwards, like a cap.

At a critical point the gradient is $0$. A cup has its lowest point there, and a cap has its highest point. This gives the test.

The second derivative test

Let $f'(c) = 0$.

• If $f''(c) < 0$, then $f$ has a local maximum at $c$.
• If $f''(c) > 0$, then $f$ has a local minimum at $c$.
• If $f''(c) = 0$, the test fails. The point may be a maximum, a minimum or neither. Use the first derivative test.

Take $f(x) = 2x^3 - 3x^2 - 12x + 5$ again. Then $f'(x) = 6(x - 2)(x + 1)$ and $f''(x) = 12x - 6$. At $x = -1$, $f'' = -18 < 0$, so it is a local maximum. At $x = 2$, $f'' = 18 > 0$, so it is a local minimum. This agrees with the first derivative test, and it needs fewer sign checks.

When the test fails

Three functions have $f'(0) = 0$ and $f''(0) = 0$:

The second derivative test cannot tell them apart. The first derivative test can: $f' = 4x^3$ changes from $-$ to $+$ for $x^4$.

Greatest and least values on a closed interval

A local maximum is the best value near a point. We often want the best value on the whole interval. The absolute (or global) maximum is the greatest value of $f$ on the interval, and the absolute minimum is the least.

The Extreme Value Theorem. If $f$ is continuous on a closed interval $[a, b]$, then $f$ has a greatest value and a least value on $[a, b]$.

To find them:

1. Find the critical points of $f$ inside $(a, b)$.
2. Calculate $f$ at each of these points and at the end points $a$ and $b$.
3. The largest number is the absolute maximum. The smallest is the absolute minimum.

For $f(x) = 2x^3 - 9x^2 + 12x$ on $[0, 3]$, we have $f'(x) = 6(x - 1)(x - 2)$. The critical points $1$ and $2$ are inside. The values are $f(0) = 0$, $f(1) = 5$, $f(2) = 4$ and $f(3) = 9$. The greatest value is $9$ at the end point $x = 3$, and the least is $0$ at the end point $x = 0$. The local maximum $5$ is not the greatest value.

Open and unbounded intervals

If the interval is open, or goes on for ever, the Extreme Value Theorem does not apply. The extreme value may not exist. For $f(x) = x^2$ on $(0, 3)$, the values get as close to $0$ and to $9$ as we like but never reach them.

If $f$ has only one critical point on an interval and it is a local minimum, it is the least value on the whole interval. Take $f(x) = x + \dfrac{9}{x}$ for $x > 0$. Then $f'(x) = 1 - \dfrac{9}{x^2} = 0$ gives $x = 3$, and $f''(x) = \dfrac{18}{x^3} > 0$. So the least value is $f(3) = 3 + 3 = 6$. There is no greatest value.

A worked example

Find the greatest and least values of $f(x) = x^3 - 6x^2 + 9x$ on $[0, 5]$.

1. The derivative. $f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$. Both critical points are inside $[0, 5]$.
2. The values. $f(0) = 0$, $f(1) = 4$, $f(3) = 0$ and $f(5) = 20$.
3. The answer. The greatest value is $20$, at $x = 5$. The least value is $0$, at both $x = 0$ and $x = 3$.

Practise with new numbers

Try it in Python

The program finds the greatest and the least value of $f(x) = 2x^3 - 9x^2 + 12x$ on $[0, 3]$ by comparing the end points and the critical points. It also prints $f''$ at each.

Common mistakes

• Using the second derivative test when $f''(c) = 0$. The test says nothing. Use the first derivative test.
• Forgetting the end points. On a closed interval, the greatest value is often at an end point.
• Including a critical point outside the interval. Only critical points inside $[a, b]$ count.
• Mixing up the signs. $f'' > 0$ gives a minimum, not a maximum.
• Calling the greatest local maximum the absolute maximum. Compare it with the end points.

Remember these

• $f'(c) = 0$ and $f''(c) < 0$: local maximum. $f''(c) > 0$: local minimum. $f''(c) = 0$: use another test.
• On $[a, b]$, compare $f$ at the end points and at the critical points inside.
• An open interval may have no greatest or least value.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Use the second derivative test on $f(x) = x^2 - 4x + 7$.

2. Does the second derivative test work for $f(x) = x^4$ at $x = 0$? What does the first derivative test say?

3. Classify the critical points of $f(x) = x^3 - 3x$ with the second derivative test.

4. Find the greatest and least values of $f(x) = x^2 - 2x$ on $[0, 3]$.

5. Why does $f(x) = x^2$ on the open interval $(0, 3)$ have no greatest value and no least value?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 6 (Application of Derivatives), the section on maxima and minima.
• OpenStax, Calculus Volume 1, Section 4.3 (Maxima and Minima) and Section 4.5 (Derivatives and the Shape of a Graph).