Concavity of A Function

Problem 175: Determine the concavity of the function $y = x^3$ on its domain.

Solution: The domain $y(x)$ is given by $(-\infty, \infty)$ . In order to find the concavity of a differentiable function, we use the second derivative. That is,

(1) $\begin{align*} y(x) &=x^3 \\ y'(x) & = 3x^2 \\ y"(x) & = 6x.\end{align*}$

We have a change of concavity at the inflection point (that is, $y"(x)=0$ ). Then we have

$y"(x) = 6x = 0 \quad \Rightarrow \quad x = 0.$

Since $f"(x)= 6x < 0$ on the interval $(-\infty, 0)$ , the function $f(x)$ is concave down on the interval $(-\infty, 0)$ .

Likewise, since $f"(x)= 6x > 0$ on the interval $(0, \infty)$ , the function $f(x)$ is concave up on the interval $(0, \infty)$ .

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