What is the extreme f(x)=x^2-8x,0<= x<= 9 ?

The extreme f(x)=x^2-8x,0<= x<= 9 is Maximum(0,0),Minimum(4,-16),Maximum(9,9)

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extreme f(x)=x^2-8x,0<= x<= 9

extreme points

f (x) = x^{2} - 8 x, 0 \leq x \leq 9

Maximum (0, 0), Minimum (4, - 16), Maximum (9, 9)

Solution steps

Find the critical points:

x = 0, x = 4, x = 9

Domain of

x^{2} - 8 x, 0 \leq x \leq 9 : 0 \leq x \leq 9

Find intervals:

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < 9

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = 4

into

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

Minimum (4, - 16)

Plug the extreme point

x = 9

into

x^{2} - 8 x \Rightarrow y = 9

Maximum (9, 9)

Maximum (0, 0), Minimum (4, - 16), Maximum (9, 9)

f (x) = 2 - 3 x, 0 \leq x \leq 1

f (x) = (\frac{1}{12} x^{3} + \frac{- 1}{2} x^{2} + 12)

y = x^{2} + 4 x - 7

f (x) = \frac{4}{3} x^{\frac{3}{2}} - \frac{1}{2} x^{2}

y = (x^{2} - 9)^{3}

What is the extreme f(x)=x^2-8x,0<= x<= 9 ?
The extreme f(x)=x^2-8x,0<= x<= 9 is Maximum(0,0),Minimum(4,-16),Maximum(9,9)