What is the extreme f(x)=-2/(x^2),0.5<= x<= 5 ?

The extreme f(x)=-2/(x^2),0.5<= x<= 5 is Minimum(0.5,-8),Maximum(5,-2/25)

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extreme f(x)=-2/(x^2),0.5<= x<= 5

extreme points

f (x) = - \frac{2}{x ^{2}}, 0.5 \leq x \leq 5

Minimum (0.5, - 8), Maximum (5, - \frac{2}{25})

Solution steps

Find the critical points:

x = 0.5, x = 5

Domain of

- \frac{2}{x ^{2}}, 0.5 \leq x \leq 5 : 0.5 \leq x \leq 5

Find intervals:

Increasing

: 0.5 < x < 5

Plug the extreme point

x = 0.5

into

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

Minimum (0.5, - 8)

Plug the extreme point

x = 5

into

- \frac{2}{x ^{2}} \Rightarrow y = - \frac{2}{25}

Maximum (5, - \frac{2}{25})

Minimum (0.5, - 8), Maximum (5, - \frac{2}{25})

f (x) = 2 - 3 x

x^{4} - 2 x^{2} - y^{2} + 3

f (t) = 9 cos (2 t)

S (X, Y) = X Y

f (x) = 2 x^{3} - 36 x^{2} + 162 x - 4

What is the extreme f(x)=-2/(x^2),0.5<= x<= 5 ?
The extreme f(x)=-2/(x^2),0.5<= x<= 5 is Minimum(0.5,-8),Maximum(5,-2/25)