What is the extreme f(x)=(x^2)/(x-8),-2<= x<=-1/4 ?

The extreme f(x)=(x^2)/(x-8),-2<= x<=-1/4 is Minimum(-2,-2/5),Maximum(-1/4 ,-1/132)

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extreme f(x)=(x^2)/(x-8),-2<= x<=-1/4

extreme points

f (x) = \frac{x ^{2}}{x - 8}, - 2 \leq x \leq - \frac{1}{4}

Minimum (- 2, - \frac{2}{5}), Maximum (- \frac{1}{4}, - \frac{1}{132})

Solution steps

Find the critical points:

x = - 2, x = - \frac{1}{4}

Domain of

\frac{x ^{2}}{x - 8}, - 2 \leq x \leq - \frac{1}{4} : - 2 \leq x \leq - \frac{1}{4}

Find intervals:

Increasing

: - 2 < x < - \frac{1}{4}

Plug the extreme point

x = - 2

into

\frac{x ^{2}}{x - 8} \Rightarrow y = - \frac{2}{5}

Minimum (- 2, - \frac{2}{5})

Plug the extreme point

x = - \frac{1}{4}

into

\frac{x ^{2}}{x - 8} \Rightarrow y = - \frac{1}{132}

Maximum (- \frac{1}{4}, - \frac{1}{132})

Minimum (- 2, - \frac{2}{5}), Maximum (- \frac{1}{4}, - \frac{1}{132})

f (x) = - 4 x^{4} + 120 x^{2} + 120

\frac{x ^{2} - 63}{x - 8}

f (x) = 9 x^{3} + 2 1^{2} + 8 x - 3

- 2 x^{2} + 8 x - 6

f (x) = 9 + 2 x^{2} - x^{4}

What is the extreme f(x)=(x^2)/(x-8),-2<= x<=-1/4 ?
The extreme f(x)=(x^2)/(x-8),-2<= x<=-1/4 is Minimum(-2,-2/5),Maximum(-1/4 ,-1/132)