What is the extreme f(x)=2x^3+3x^2-72x+4,-4<= x<= 6 ?

The extreme f(x)=2x^3+3x^2-72x+4,-4<= x<= 6 is Maximum(-4,212),Minimum(3,-131),Maximum(6,112)

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extreme f(x)=2x^3+3x^2-72x+4,-4<= x<= 6

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 72 x + 4, - 4 \leq x \leq 6

Maximum (- 4, 212), Minimum (3, - 131), Maximum (6, 112)

Solution steps

Find the critical points:

x = - 4, x = 3, x = 6

Domain of

2 x^{3} + 3 x^{2} - 72 x + 4, - 4 \leq x \leq 6 : - 4 \leq x \leq 6

Find intervals:

Decreasing

: - 4 < x < 3,

Increasing

: 3 < x < 6

Plug the extreme point

x = - 4

into

2 x^{3} + 3 x^{2} - 72 x + 4 \Rightarrow y = 212

Maximum (- 4, 212)

Plug the extreme point

x = 3

into

2 x^{3} + 3 x^{2} - 72 x + 4 \Rightarrow y = - 131

Minimum (3, - 131)

Plug the extreme point

x = 6

into

2 x^{3} + 3 x^{2} - 72 x + 4 \Rightarrow y = 112

Maximum (6, 112)

Maximum (- 4, 212), Minimum (3, - 131), Maximum (6, 112)

2 x

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

f (x, y) = 3 x^{3} - 12 x + 2 y^{2} - 8 y + 7

f (x, y) = (\frac{1}{16 - x ^{2} - y ^{2}})

6 x

What is the extreme f(x)=2x^3+3x^2-72x+4,-4<= x<= 6 ?
The extreme f(x)=2x^3+3x^2-72x+4,-4<= x<= 6 is Maximum(-4,212),Minimum(3,-131),Maximum(6,112)