What is the extreme f(x)=3x^2+8,-4<= x<= 4 ?

The extreme f(x)=3x^2+8,-4<= x<= 4 is Maximum(-4,56),Minimum(0,8),Maximum(4,56)

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extreme f(x)=3x^2+8,-4<= x<= 4

extreme points

f (x) = 3 x^{2} + 8, - 4 \leq x \leq 4

Maximum (- 4, 56), Minimum (0, 8), Maximum (4, 56)

Solution steps

Find the critical points:

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

Domain of

3 x^{2} + 8, - 4 \leq x \leq 4 : - 4 \leq x \leq 4

Find intervals:

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < 4

Plug the extreme point

x = - 4

into

3 x^{2} + 8 \Rightarrow y = 56

Maximum (- 4, 56)

Plug the extreme point

x = 0

into

3 x^{2} + 8 \Rightarrow y = 8

Minimum (0, 8)

Plug the extreme point

x = 4

into

3 x^{2} + 8 \Rightarrow y = 56

Maximum (4, 56)

Maximum (- 4, 56), Minimum (0, 8), Maximum (4, 56)

f (x) = x (\frac{3600}{x - 4} + 2 \cdot 1 0^{- 3})

f (x) = x^{2} - 18 ln (x)

f (x, y) = 5 x^{2} + 5 y^{2} + 20 x - 10 y

f (x, y) = x y + \frac{π}{8} x^{2}

f (x) = - 4 + x - x^{2}

What is the extreme f(x)=3x^2+8,-4<= x<= 4 ?
The extreme f(x)=3x^2+8,-4<= x<= 4 is Maximum(-4,56),Minimum(0,8),Maximum(4,56)