7.4 The second derivative, convexity and points of inflection — Practise

Work through these on your own. Show the first derivative, then the second. For every candidate point of inflection, test the sign of the second derivative on both sides before you decide.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find $f''(x)$ for $f(x) = x^4 - 6x^2$ .
Find the second derivative of $y = 3x^5$ .
Find $\dfrac{d^2y}{dx^2}$ for $y = \sin 2x$ .
Find the second derivative of $y = e^{-x}$ .
For $f(x) = x^3 - 3x^2 + 2x$ , find the value of $x$ where $f''(x) = 0$ .
The curve $y = x^2 - 4x + 1$ has a turning point at $x = 2$ . Use the second derivative to classify it.
Find the second derivative of $f(x) = 3x^2 - 6x + 1$ , and state whether the curve is convex or concave everywhere.
For $f(x) = x^3 - 6x^2$ , show that $f''(x) = 0$ at $x = 2$ and that the second derivative changes sign there.
For $f(x) = x^4$ , show that $f''(0) = 0$ but the second derivative does not change sign at $x = 0$ .
Find the second derivative of $y = \ln x$ for $x > 0$ .

Answers and concise solutions

1. The second derivative

Answer: $f'(x) = 4x^3 - 12x$ , so $f''(x) = 12x^2 - 12$ .

2. The second derivative

Answer: First derivative $= 15x^4$ , so second derivative $= 60x^3$ .

3. The second derivative

Answer: $\dfrac{dy}{dx} = 2\cos 2x$ , so $\dfrac{d^2y}{dx^2} = -4\sin 2x$ . The minus sign returns after two differentiations.

4. The second derivative

Answer: $\dfrac{dy}{dx} = -e^{-x}$ , so $\dfrac{d^2y}{dx^2} = e^{-x}$ . Each differentiation multiplies by $-1$ , and two minus signs make a plus.

5. The value of $x$

Answer: $f'(x) = 3x^2 - 6x + 2$ and $f''(x) = 6x - 6$ . Setting this to zero gives $6x = 6$ , so $x = 1$ .

6. The classification

Answer: The second derivative is $f''(x) = 2$ , which is positive, so the curve is convex. The turning point is a minimum.

7. The second derivative

Answer: The second derivative is the constant $6$ , which is positive for every $x$ . The curve is convex everywhere.

8. The two signs

Answer: $f'(x) = 3x^2 - 12x$ and $f''(x) = 6x - 12$ , which is zero at $x = 2$ . At $x = 1$ it is $-6 < 0$ (concave); at $x = 3$ it is $6 > 0$ (convex). The sign changes, so $x = 2$ is a point of inflection.

9. The two signs

Answer: $f''(x) = 12x^2$ , so $f''(0) = 0$ . For $x = -1$ and for $x = 1$ the value is $12 > 0$ , so the curve is convex on both sides. The second derivative does not change sign, so $x = 0$ is not a point of inflection.

10. The second derivative

Answer: $\dfrac{dy}{dx} = \dfrac{1}{x} = x^{-1}$ , so $\dfrac{d^2y}{dx^2} = -x^{-2} = -\dfrac{1}{x^2}$ . The second derivative is always negative, so the curve is concave.