What is the extreme f(x)=8(x^2)/((x-6)),-3<= x<= 2 ?

The extreme f(x)=8(x^2)/((x-6)),-3<= x<= 2 is Minimum(-3,-8),Maximum(0,0),Minimum(2,-8)

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extreme f(x)=8(x^2)/((x-6)),-3<= x<= 2

extreme points

f (x) = 8 \frac{x ^{2}}{( x - 6 )}, - 3 \leq x \leq 2

Minimum (- 3, - 8), Maximum (0, 0), Minimum (2, - 8)

Solution steps

Find the critical points:

x = - 3, x = 0, x = 2

Domain of

8 \frac{x ^{2}}{x - 6}, - 3 \leq x \leq 2 : - 3 \leq x \leq 2

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 2

Plug the extreme point

x = - 3

into

8 \frac{x ^{2}}{x - 6} \Rightarrow y = - 8

Minimum (- 3, - 8)

Plug the extreme point

x = 0

into

8 \frac{x ^{2}}{x - 6} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 2

into

8 \frac{x ^{2}}{x - 6} \Rightarrow y = - 8

Minimum (2, - 8)

Minimum (- 3, - 8), Maximum (0, 0), Minimum (2, - 8)

f (x) = - 0.05 x^{2} + 300 x + 2000

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

f (x) = - x^{(3)} + 12 x^{(2)} + 45 x - 52

f (x) = 2 cos (x) + x

f (x) = 2 cos (x) - 1

What is the extreme f(x)=8(x^2)/((x-6)),-3<= x<= 2 ?
The extreme f(x)=8(x^2)/((x-6)),-3<= x<= 2 is Minimum(-3,-8),Maximum(0,0),Minimum(2,-8)