The Second Derivative and the Shape of a Function f(x)

Practice 3: Which of the labelled points in Fig. 8 are inflection points?

Fig. 8

To find the inflection points of a function we can use the second derivative of the function. If $f^{\prime \prime}(x) > 0$, then the graph of $f$ is concave up at the point $(x, f(x))$ so $(x, f(x))$ is not an inflection point. Similarly, if $f^{\prime \prime}(x) < 0$, then the graph of $\mathrm{f}$ is concave down at the point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$ and the point is not an inflection point. The only points left which can possibly be inflection points are the places where $\mathrm{f}^{\prime \prime}(\mathrm{x})$ is $0$ or undefined ($f'$ is not differentiable). To find the inflection points of a function we only need to check the points where $\mathrm{f}^{\prime \prime}(\mathrm{x})$ is $0$ or undefined. If $\mathrm{f}^{\prime \prime}(\mathrm{c})=0$ or is undefined, then the point $(\mathrm{c}, \mathrm{f}(\mathrm{c}))$ may or may not be an inflection point – we would need to check the concavity of $\mathrm{f}$ on each side of $\mathrm{x}=\mathrm{c}$. The functions in the next example illustrate what can happen.

Example 2: Let $f(x)=x^{3}, g(x)=x^{4}$ and $h(x)=x^{1 / 3}$ (Fig. 9). For which of these functions is the point $(0,0)$ an inflection point?

Fig. 9

Solution: Graphically, it is clear that the concavity of $f(x)=x^{3}$ and $\mathrm{h}(\mathrm{x})=\mathrm{x}^{1 / 3}$ changes at $(0,0)$, so $(0,0)$ is an inflection point for $\mathrm{f}$ and $\mathrm{h}$. The function $\mathrm{g}(\mathrm{x})=\mathrm{x}^{4}$ is concave up everywhere so $(0,0)$ is not an inflection point of $\mathrm{g}$.

If $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$ and $f^{\prime \prime}(x)=6 x$. The only point at which $f^{\prime \prime}(x)=0$ or is undefined ($f'$ is not differentiable) is at $x=0$. If $x < 0$, then $f^{\prime \prime}(x) < 0$ so $f$ is concave down. If $x > 0$, then $\mathrm{f}^{\prime \prime}(\mathrm{x}) > 0$ so $\mathrm{f}$ is concave up. At $\mathrm{x}=0$ the concavity changes so the point $(0, \mathrm{f}(0))=(0,0)$ is an inflection point of $\mathrm{x}^{3}$.

If $\mathrm{g}(\mathrm{x})=\mathrm{x}^{4}$, then $\mathrm{g}^{\prime}(\mathrm{x})=4 \mathrm{x}^{3}$ and $\mathrm{g}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}^{2}$. The only point at which $\mathrm{g}^{\prime \prime}(\mathrm{x})=0$ or is undefined is at $x=0$. If $x < 0$, then $g^{\prime \prime}(x) > 0$ so $g$ is concave up. If $x > 0$, then $g^{\prime \prime}(x) > 0$ so $g$ is also concave up. At $x=0$ the concavity does not change so the point $(0, g(0))=(0,0)$ is not an inflection point of $x^{4}$.

If $\mathrm{h}(\mathrm{x})=\mathrm{x}^{1 / 3}$, then $\mathrm{h}^{\prime}(\mathrm{x})=\frac{1}{3} \mathrm{x}^{-2 / 3}$ and $\mathrm{h}^{\prime \prime} (\mathrm{x})=-\frac{2}{9} \mathrm{x}^{-5 / 3} \cdot \mathrm{h}^{\prime \prime}$ is not defined if $\mathrm{x}=0$, but $\mathrm{h}^{\prime \prime}$ (negative number) $> 0$ and $\mathrm{h}^{\prime \prime}$ (positive number) $< 0$ so $\mathrm{h}$ changes concavity at $(0,0)$ and $(0,0)$ is an inflection point of $\mathrm{h}$.

Practice 4: Find the inflection points of $f(x)=x^{4}-12 x^{3}+30 x^{2}+5 x-7$.

Example 3: Sketch graph of a function with $f(2)=3, f^{\prime}(2)=1$, and an inflection point at $(2,3)$. Solution: Two solutions are given in Fig. 10.

Fig. 10