The Shape of a Graph, Part I — Question 1

Question 1 - Solution

We are given:

$$f(x) = x^4 - 4x^3 + 6x^2$$

Step 1: Find the second derivative

$$\begin{aligned} f'(x) &= 4x^3 - 12x^2 + 12x \\ f''(x) &= 12x^2 - 24x + 12 = 12(x - 1)^2 \end{aligned}$$

(a) Since $f''(x) = 12(x - 1)^2 \ge 0$, the second derivative is never negative.

For all $x \neq 1$, $f''(x) > 0$, so the function is concave up.

At $x = 1$, the second derivative is zero, but the sign of $f''(x)$ does not change.

Answer

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

Concave down: $\boxed{\text{None}}$

(b) Because the concavity does not change at $x = 1$, there is no inflection point.

Answer

$\boxed{\text{No inflection points}}$

Graph of the function