What is the extreme f(x)=2x^3+3x^2-72x+1,-4<= x<= 7 ?

The extreme f(x)=2x^3+3x^2-72x+1,-4<= x<= 7 is Maximum(-4,209),Minimum(3,-134),Maximum(7,330)

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extreme f(x)=2x^3+3x^2-72x+1,-4<= x<= 7

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 72 x + 1, - 4 \leq x \leq 7

Maximum (- 4, 209), Minimum (3, - 134), Maximum (7, 330)

Solution steps

Find the critical points:

x = - 4, x = 3, x = 7

Domain of

2 x^{3} + 3 x^{2} - 72 x + 1, - 4 \leq x \leq 7 : - 4 \leq x \leq 7

Find intervals:

Decreasing

: - 4 < x < 3,

Increasing

: 3 < x < 7

Plug the extreme point

x = - 4

into

2 x^{3} + 3 x^{2} - 72 x + 1 \Rightarrow y = 209

Maximum (- 4, 209)

Plug the extreme point

x = 3

into

2 x^{3} + 3 x^{2} - 72 x + 1 \Rightarrow y = - 134

Minimum (3, - 134)

Plug the extreme point

x = 7

into

2 x^{3} + 3 x^{2} - 72 x + 1 \Rightarrow y = 330

Maximum (7, 330)

Maximum (- 4, 209), Minimum (3, - 134), Maximum (7, 330)

x^{3} + 3 x^{2} + 9 x

f (x) = \frac{( x ^{2} )}{x ^{2} + 243}

f (x) = sec^{2} (x) - 23 tan (x)

f (x) = 4 x^{3} - 12 x^{2} - 96 x + 9

f (x) = 4 x^{3} - 12 x^{2} - 96 x + 7

What is the extreme f(x)=2x^3+3x^2-72x+1,-4<= x<= 7 ?
The extreme f(x)=2x^3+3x^2-72x+1,-4<= x<= 7 is Maximum(-4,209),Minimum(3,-134),Maximum(7,330)