What is the extreme f(x)=4x^2+40,-1<= x<= 1 ?

The extreme f(x)=4x^2+40,-1<= x<= 1 is Maximum(-1,44),Minimum(0,40),Maximum(1,44)

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extreme f(x)=4x^2+40,-1<= x<= 1

extreme points

f (x) = 4 x^{2} + 40, - 1 \leq x \leq 1

Maximum (- 1, 44), Minimum (0, 40), Maximum (1, 44)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1

Domain of

4 x^{2} + 40, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

4 x^{2} + 40 \Rightarrow y = 44

Maximum (- 1, 44)

Plug the extreme point

x = 0

into

4 x^{2} + 40 \Rightarrow y = 40

Minimum (0, 40)

Plug the extreme point

x = 1

into

4 x^{2} + 40 \Rightarrow y = 44

Maximum (1, 44)

Maximum (- 1, 44), Minimum (0, 40), Maximum (1, 44)

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

y = x 64 - x^{2}

f (x) = \frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x + 2

f (x) = - 3 x^{2} + 3 x y - 2 x + y^{2} + 6 y + 12

f (x) = \frac{x}{16 + x ^{2}}

What is the extreme f(x)=4x^2+40,-1<= x<= 1 ?
The extreme f(x)=4x^2+40,-1<= x<= 1 is Maximum(-1,44),Minimum(0,40),Maximum(1,44)