Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 − 9x2 − 108x + 6, [−3, 4]

Answer 1

Answer:

The absolute maximum of f(x) on [-3, 4] is 138 and the absolute minimum of f(x) on [-3, 4] is -237.

Step-by-step explanation:

To find the absolute extrema values of $f(x) = 6x^3 - 9x^2 - 108x + 6$  on the closed interval [−3, 4] you must:

1. Locate all critical values. We need to find the derivative of the function and set it equal to zero.

$\frac{d}{dx}f(x)= \frac{d}{dx}\left(6x^3-9x^2-108x+6\right)=\\\\f'(x)=\frac{d}{dx}\left(6x^3\right)-\frac{d}{dx}\left(9x^2\right)-\frac{d}{dx}\left(108x\right)+\frac{d}{dx}\left(6\right)\\\\f'(x)=18x^2-18x-108$

$18x^2-18x-108=0\\18\left(x^2-x-6\right)=0\\18\left(x+2\right)\left(x-3\right)=0\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\x=-2,\:x=3$

2. Evaluate f(x) at all the critical values and also at the two values -3 and 4

$\left\begin{array}{cc}x&f(x)\\-3&87\\-2&138\\3&-237\\4&-186\end{array}\right$

3. The absolute maximum of f(x) on [-3, 4] will be the largest number found in Step 2, while the absolute minimum of f(x) on [-3, 4] will be the smallest number found in Step 2.

Therefore,

The absolute maximum of f(x) on [-3, 4] is 138 and the absolute minimum of f(x) on [-3, 4] is -237.

Answer 2

It is convenient to use a graphing calculator for this. The graph shows the maximum and minimum are not at the ends of the interval, so could be found by differentiating the function and setting that derivative to zero. The function would then need to be evaluated for those solutions {-2, 3}.

The absolute minimum on the interval is -237 at x=3.
The absolute maximum on the interval is 138 at x=-2.