What is the extreme x^3-3/2 x^2,-1<= x<= 4 ?

The extreme x^3-3/2 x^2,-1<= x<= 4 is Minimum(-1,-5/2),Maximum(0,0),Minimum(1,-1/2),Maximum(4,40)

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extreme x^3-3/2 x^2,-1<= x<= 4

extreme points

x^{3} - \frac{3}{2} x^{2}, - 1 \leq x \leq 4

Minimum (- 1, - \frac{5}{2}), Maximum (0, 0), Minimum (1, - \frac{1}{2}), Maximum (4, 40)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1, x = 4

Domain of

x^{3} - \frac{3}{2} x^{2}, - 1 \leq x \leq 4 : - 1 \leq x \leq 4

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 4

Plug the extreme point

x = - 1

into

x^{3} - \frac{3}{2} x^{2} \Rightarrow y = - \frac{5}{2}

Minimum (- 1, - \frac{5}{2})

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = 1

into

x^{3} - \frac{3}{2} x^{2} \Rightarrow y = - \frac{1}{2}

Minimum (1, - \frac{1}{2})

Plug the extreme point

x = 4

into

x^{3} - \frac{3}{2} x^{2} \Rightarrow y = 40

Maximum (4, 40)

Minimum (- 1, - \frac{5}{2}), Maximum (0, 0), Minimum (1, - \frac{1}{2}), Maximum (4, 40)

f (x) = ln (x^{2} + 3 x + 12)

f (x) = x^{3} - 6 x^{2} + 9 x + 7

1 - \frac{2}{x ^{2} + 1}

g (x) = 2 x^{3} - 6 x + 3, - 2 \leq x \leq 2

x (\frac{200 - x}{2})

What is the extreme x^3-3/2 x^2,-1<= x<= 4 ?
The extreme x^3-3/2 x^2,-1<= x<= 4 is Minimum(-1,-5/2),Maximum(0,0),Minimum(1,-1/2),Maximum(4,40)