The Shape of a Graph, Part II — Question 4

Question 4 - Solution

We are given $$f(x) = x^3 - 3x$$

First derivative

$$f'(x) = 3x^2 - 3$$

Second derivative

$$f''(x) = 6x$$

(a) Concavity

Setting $f''(x) = 0$ gives $$x = 0.$$

For $x < 0$, the second derivative is negative, so the graph is concave down.

For $x > 0$, the second derivative is positive, so the graph is concave up.

Concave down on $$\boxed{(-\infty, 0)}$$

Concave up on $$\boxed{(0, \infty)}$$

(b) Inflection point

Because the concavity changes at $x = 0$, an inflection point occurs there.

$$f(0) = 0$$

Inflection point at $$\boxed{(0, 0)}$$

Graph of $f(x)$