The Shape of a Graph, Part II — Question 7

Question 7

Let $$f(x) = x^3 - 6x^2 + 12x - 5$$

(a) Determine the intervals where the graph of $f$ is concave up and concave down.

(b) Find all inflection points.

Question 7 - Solution

We are given $$f(x) = x^3 - 6x^2 + 12x - 5$$

First derivative

$$f'(x) = 3x^2 - 12x + 12$$

Second derivative

$$f''(x) = 6x - 12$$

(a) Concavity

Setting $f''(x) = 0$ gives $$x = 2.$$

For $x < 2$, the second derivative is negative, so the graph is concave down.

For $x > 2$, the second derivative is positive, so the graph is concave up.

Concave down on $$\boxed{(-\infty, 2)}$$

Concave up on $$\boxed{(2, \infty)}$$

(b) Inflection point

Because the concavity changes at $x = 2$, an inflection point occurs there.

$$f(2) = 3$$

Inflection point at $$\boxed{(2, 3)}$$

Graph of $f(x)$