What is the extreme f(x)=x^2+3,-1<= x<= 4 ?

The extreme f(x)=x^2+3,-1<= x<= 4 is Maximum(-1,4),Minimum(0,3),Maximum(4,19)

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extreme f(x)=x^2+3,-1<= x<= 4

extreme points

f (x) = x^{2} + 3, - 1 \leq x \leq 4

Maximum (- 1, 4), Minimum (0, 3), Maximum (4, 19)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 4

Domain of

x^{2} + 3, - 1 \leq x \leq 4 : - 1 \leq x \leq 4

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 4

Plug the extreme point

x = - 1

into

x^{2} + 3 \Rightarrow y = 4

Maximum (- 1, 4)

Plug the extreme point

x = 0

into

x^{2} + 3 \Rightarrow y = 3

Minimum (0, 3)

Plug the extreme point

x = 4

into

x^{2} + 3 \Rightarrow y = 19

Maximum (4, 19)

Maximum (- 1, 4), Minimum (0, 3), Maximum (4, 19)

f (x, y) = 5 x + 5 y

f (x) = \frac{1}{4} (- 4 x^{3} + 2 x^{2} + 12)

f (x, y) = x^{2} - 2 y^{2} + 8 x - 3 y + 1

f (2) = \frac{x}{e ^{- x}}

f (x) = 5 sin (x) cos (x), (- π, π)

What is the extreme f(x)=x^2+3,-1<= x<= 4 ?
The extreme f(x)=x^2+3,-1<= x<= 4 is Maximum(-1,4),Minimum(0,3),Maximum(4,19)