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)

Popular >

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

extreme points

f (x) = x^{4} - 14 x^{2} + 49, 0 \leq x \leq 5

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)

f (x) = ln (x^{2} + 5 x + 9), - 3 \leq x \leq 3

x^{3} - 3 x^{2} - 24 x + 8

f (x) = t^{3} - 9 t^{2}

y = \frac{( x ( z + 2 ) )}{7}

- \frac{x ^{4}}{4} + \frac{x ^{2}}{4}

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)