What is the extreme f(x)=(x-1)^2,0<= x<= 4 ?

The extreme f(x)=(x-1)^2,0<= x<= 4 is Maximum(0,1),Minimum(1,0),Maximum(4,9)

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extreme f(x)=(x-1)^2,0<= x<= 4

extreme points

f (x) = (x - 1)^{2}, 0 \leq x \leq 4

Maximum (0, 1), Minimum (1, 0), Maximum (4, 9)

Solution steps

Find the critical points:

x = 0, x = 1, x = 4

Domain of

(x - 1)^{2}, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 4

Plug the extreme point

x = 0

into

(x - 1)^{2} \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 1

into

(x - 1)^{2} \Rightarrow y = 0

Minimum (1, 0)

Plug the extreme point

x = 4

into

(x - 1)^{2} \Rightarrow y = 9

Maximum (4, 9)

Maximum (0, 1), Minimum (1, 0), Maximum (4, 9)

(3 - 2) (37)

f (x) = \frac{( sin ( 5 x ) )}{1 + sin ( 5 x )}

1 - 6^{t}

\frac{x ^{2} + x + 1}{x}

x = 6 y + 5

What is the extreme f(x)=(x-1)^2,0<= x<= 4 ?
The extreme f(x)=(x-1)^2,0<= x<= 4 is Maximum(0,1),Minimum(1,0),Maximum(4,9)