First slide

Maxima and minima

Question

$The image of the interval [- 1, 3] under the maping f (x) = 4 x^{3} - 12 x is$

Moderate

A

$[- 2, 0]$
B

$[- 8, 72]$
C

$[- 8, 0]$
D

$none of these$

Solution

To find the image of the given interval, we must find the set
of values of $f (x) f o r x \in [- 1, 3]$ . By virtue of the continuity
of f(x) , the image is the interval $[\min_{x \in [- 1, 3]} f (x), \max_{x \in [- 1, 3]} f (x)]$
$\begin{array}{l} The critical points of f (x) are given by \\ f^{'} (x) = 12 x^{2} - 12 = 12 (x^{2} - 1) = 0 . That is, \\ x = \pm 1, sothat f (1) = 4 \cdot 1 - 12 = - 8, f (- 1) = - 4 + 12 = 8 \\ and f (3) = 4 \cdot 27 - 12 \cdot 3 = 108 - 36 = 72 \\ ∴ max_{x \in [- 1, 3]} f (x) = f (3) = 72 \end{array}$
$\begin{array}{l} And min_{x \in [- 1, 3]} f (x) = f (1) = - 8 \\ Hence the image of [- 1, 3] under the mapping \\ f (x) is [- 8, 72] \end{array}$

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