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

The extreme f(x)= 1/x-2/(x^2),-2<= x<= 1 is Maximum(-2,-1),Maximum(1,-1)

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

extreme points

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

Maximum (- 2, - 1), Maximum (1, - 1)

Solution steps

Find the critical points:

x = - 2, x = 1

Domain of

\frac{1}{x} - \frac{2}{x ^{2}}, - 2 \leq x \leq 1 : - 2 \leq x < 0 or 0 < x \leq 1

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

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

Maximum (- 2, - 1)

Plug the extreme point

x = 1

into

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

Maximum (1, - 1)

Maximum (- 2, - 1), Maximum (1, - 1)

f (x) = (x^{2} - 2 x + 2) e^{1 - x}

y = \frac{x ^{3}}{- x ^{2} + 4}

4 x^{3} + 96 x^{2}

f (x) = x^{2} + 8 x + 15

(x - 2) (x^{2} - 4 x - 8)

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