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 [tex]f(x) = 6x^3 - 9x^2 - 108x + 6[/tex] 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.
[tex]\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[/tex]
[tex]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[/tex]
2. Evaluate f(x) at all the critical values and also at the two values -3 and 4
[tex]\left\begin{array}{cc}x&f(x)\\-3&87\\-2&138\\3&-237\\4&-186\end{array}\right[/tex]
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.
