What is the extreme f(x)=2x^3+3x^2-120x,0<= x<= 5 ?

The extreme f(x)=2x^3+3x^2-120x,0<= x<= 5 is Maximum(0,0),Minimum(4,-304),Maximum(5,-275)

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extreme f(x)=2x^3+3x^2-120x,0<= x<= 5

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 120 x, 0 \leq x \leq 5

Maximum (0, 0), Minimum (4, - 304), Maximum (5, - 275)

Solution steps

Find the critical points:

x = 0, x = 4, x = 5

Domain of

2 x^{3} + 3 x^{2} - 120 x, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < 5

Plug the extreme point

x = 0

into

2 x^{3} + 3 x^{2} - 120 x \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 4

into

2 x^{3} + 3 x^{2} - 120 x \Rightarrow y = - 304

Minimum (4, - 304)

Plug the extreme point

x = 5

into

2 x^{3} + 3 x^{2} - 120 x \Rightarrow y = - 275

Maximum (5, - 275)

Maximum (0, 0), Minimum (4, - 304), Maximum (5, - 275)

f (x) = 2 x^{3} - 6 x + 3

3 x - x^{3}

f (x, y) = y ln (- y + x)

f (x) = - 2 x^{3} - 9 x^{2} + 24 x + 40

f (x, y) = x^{2} y + y^{3} - 48 y

What is the extreme f(x)=2x^3+3x^2-120x,0<= x<= 5 ?
The extreme f(x)=2x^3+3x^2-120x,0<= x<= 5 is Maximum(0,0),Minimum(4,-304),Maximum(5,-275)