Review question

# When is $f(x) = 2x^3 − 9x^2 + 12x + 3$ smallest over this range? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6868

## Solution

The smallest value of the function $f(x) = 2x^3 − 9x^2 + 12x + 3$ in the range $0 \leq x \leq 2$ is

1. $1$,

2. $3$,

3. $5$,

4. $7$.

The smallest value will occur either at a stationary point of $f(x)$, or at one of the endpoints of the range, $x = 0$ or $x = 2$.

To find stationary points, we solve $f'(x) = 6x^2 - 18x + 12 = 0$, or equivalently $0 =x^2 - 3x + 2 = (x - 1)(x - 2)$.

So $x = 1$ and $x = 2$ are the stationary points of $f(x)$, and are in the desired range. From our knowledge of the shape of cubics, or from looking at $f''(x)$, we know that $x = 1$ is a maximum and $x = 2$ is a minimum.

Then the minimum value of $f(x)$ is either at $x = 2$ or $x = 0$, so we can check the values. We have \begin{align*} f(2) &= 16 - 36 + 24 + 3 = 7 \\ f(0) &= 3. \end{align*}

So the minimum value of $f(x)$ for $0 \leq x \leq 2$ is $3$, and the answer is (b).