What is the extreme f(x)=2x^3+3x^2-120x+4,-5<= x<= 5 ?

The extreme f(x)=2x^3+3x^2-120x+4,-5<= x<= 5 is Maximum(-5,429),Minimum(4,-300),Maximum(5,-271)

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extreme f(x)=2x^3+3x^2-120x+4,-5<= x<= 5

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 120 x + 4, - 5 \leq x \leq 5

Maximum (- 5, 429), Minimum (4, - 300), Maximum (5, - 271)

Solution steps

Find the critical points:

x = - 5, x = 4, x = 5

Domain of

2 x^{3} + 3 x^{2} - 120 x + 4, - 5 \leq x \leq 5 : - 5 \leq x \leq 5

Find intervals:

Decreasing

: - 5 < x < 4,

Increasing

: 4 < x < 5

Plug the extreme point

x = - 5

into

2 x^{3} + 3 x^{2} - 120 x + 4 \Rightarrow y = 429

Maximum (- 5, 429)

Plug the extreme point

x = 4

into

2 x^{3} + 3 x^{2} - 120 x + 4 \Rightarrow y = - 300

Minimum (4, - 300)

Plug the extreme point

x = 5

into

2 x^{3} + 3 x^{2} - 120 x + 4 \Rightarrow y = - 271

Maximum (5, - 271)

Maximum (- 5, 429), Minimum (4, - 300), Maximum (5, - 271)

y = 4 (1.5)^{x}

x - 4 - 4

f (x) = 6 - (8 + 5 x)^{\frac{2}{5}}

f (x, y) = 3 x^{2} + 2 y^{2} - 4 y

f (x) = \frac{3}{7} (x^{2} - 25)^{\frac{2}{3}}

What is the extreme f(x)=2x^3+3x^2-120x+4,-5<= x<= 5 ?
The extreme f(x)=2x^3+3x^2-120x+4,-5<= x<= 5 is Maximum(-5,429),Minimum(4,-300),Maximum(5,-271)