What is the extreme f(x)=x^4-14x^2+49,0<= x<= 5 ?

The extreme f(x)=x^4-14x^2+49,0<= x<= 5 is Maximum(0,49),Minimum(sqrt(7),0),Maximum(5,324)

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extreme f(x)=x^4-14x^2+49,0<= x<= 5

extreme

Maximum (0, 49), Minimum (7, 0), Maximum (5, 324)

Solution steps

Find the critical points:

x = 0, x = 7, x = 5

Domain of

x^{4} - 14 x^{2} + 49, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 7,

Increasing

: 7 < x < 5

Plug the extreme point

x = 0

into

x^{4} - 14 x^{2} + 49 \Rightarrow y = 49

Maximum (0, 49)

Plug the extreme point

x = 7

into

x^{4} - 14 x^{2} + 49 \Rightarrow y = 0

Minimum (7, 0)

Plug the extreme point

x = 5

into

x^{4} - 14 x^{2} + 49 \Rightarrow y = 324

Maximum (5, 324)

Maximum (0, 49), Minimum (7, 0), Maximum (5, 324)

What is the extreme f(x)=x^4-14x^2+49,0<= x<= 5 ?
The extreme f(x)=x^4-14x^2+49,0<= x<= 5 is Maximum(0,49),Minimum(sqrt(7),0),Maximum(5,324)