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

The extreme f(x)=x^2-8x,-infinity <x<= 8 is Minimum(4,-16),Maximum(8,0)

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extreme f(x)=x^2-8x,-infinity <x<= 8

extreme points

f (x) = x^{2} - 8 x, - \infty < x \leq 8

Minimum (4, - 16), Maximum (8, 0)

Solution steps

Find the critical points:

x = 4, x = 8

Domain of

x^{2} - 8 x, - \infty < x \leq 8 : x \leq 8

Find intervals:

Decreasing

: - \infty < x < 4,

Increasing

: 4 < x < 8

Plug the extreme point

x = 4

into

x^{2} - 8 x \Rightarrow y = - 16

Minimum (4, - 16)

Plug the extreme point

x = 8

into

x^{2} - 8 x \Rightarrow y = 0

Maximum (8, 0)

Minimum (4, - 16), Maximum (8, 0)

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What is the extreme f(x)=x^2-8x,-infinity <x<= 8 ?
The extreme f(x)=x^2-8x,-infinity <x<= 8 is Minimum(4,-16),Maximum(8,0)