When is $f(x) = 2x^3 − 9x^2 + 12x + 3$ smallest over this range?

When is $f(x) = 2x^3 − 9x^2 + 12x + 3$ smallest over this range?

Question

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

1. $1$,

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1. $3$,

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1. $5$,

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1. $7$.

Suggestion

Could we see ‘the smallest value’ as ‘the minimum value’?

How do we find the minimum values of a function?

Could we sketch the graph of $y = f(x)$?

Solution

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).