What is the extreme f(x)=2x^3-15x^2+24x,0<= x<= 5 ?

The extreme f(x)=2x^3-15x^2+24x,0<= x<= 5 is Minimum(0,0),Maximum(1,11),Minimum(4,-16),Maximum(5,-5)

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extreme f(x)=2x^3-15x^2+24x,0<= x<= 5

extreme points

f (x) = 2 x^{3} - 15 x^{2} + 24 x, 0 \leq x \leq 5

Minimum (0, 0), Maximum (1, 11), Minimum (4, - 16), Maximum (5, - 5)

Solution steps

Find the critical points:

x = 0, x = 1, x = 4, x = 5

Domain of

2 x^{3} - 15 x^{2} + 24 x, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 4,

Increasing

: 4 < x < 5

Plug the extreme point

x = 0

into

2 x^{3} - 15 x^{2} + 24 x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

2 x^{3} - 15 x^{2} + 24 x \Rightarrow y = 11

Maximum (1, 11)

Plug the extreme point

x = 4

into

2 x^{3} - 15 x^{2} + 24 x \Rightarrow y = - 16

Minimum (4, - 16)

Plug the extreme point

x = 5

into

2 x^{3} - 15 x^{2} + 24 x \Rightarrow y = - 5

Maximum (5, - 5)

Minimum (0, 0), Maximum (1, 11), Minimum (4, - 16), Maximum (5, - 5)

y = \frac{x}{x ^{2} + 64}

f (x) = 3 x^{4} - 324 x, 0 \leq x \leq 4

f (x) = - x^{2} + 1, - 1 \leq x \leq 0

f (x) = x^{5} - x^{3} + 8, - 1 \leq x \leq 1

f (x) = 2 x^{2} - 8 x + 2

What is the extreme f(x)=2x^3-15x^2+24x,0<= x<= 5 ?
The extreme f(x)=2x^3-15x^2+24x,0<= x<= 5 is Minimum(0,0),Maximum(1,11),Minimum(4,-16),Maximum(5,-5)