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

The extreme f(x)=x^4-2x^3,-2<= x<= 2 is Maximum(-2,32),Saddle(0,0),Minimum(3/2 ,-27/16),Maximum(2,0)

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

extreme points

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

Maximum (- 2, 32), Saddle (0, 0), Minimum (\frac{3}{2}, - \frac{27}{16}), Maximum (2, 0)

Solution steps

Find the critical points:

x = - 2, x = 0, x = \frac{3}{2}, x = 2

Domain of

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

Find intervals:

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < 2

Plug the extreme point

x = - 2

into

x^{4} - 2 x^{3} \Rightarrow y = 32

Maximum (- 2, 32)

Plug the extreme point

x = 0

into

x^{4} - 2 x^{3} \Rightarrow y = 0

Saddle (0, 0)

Plug the extreme point

x = \frac{3}{2}

into

x^{4} - 2 x^{3} \Rightarrow y = - \frac{27}{16}

Minimum (\frac{3}{2}, - \frac{27}{16})

Plug the extreme point

x = 2

into

x^{4} - 2 x^{3} \Rightarrow y = 0

Maximum (2, 0)

Maximum (- 2, 32), Saddle (0, 0), Minimum (\frac{3}{2}, - \frac{27}{16}), Maximum (2, 0)

f (x) = \frac{7 x}{x ^{2} - 49}

f (x, y) = 9 x^{3} + \frac{y ^{3}}{3} - 4 x y

f (x, y) = 2 x y^{4} - 5 x^{2} y^{3} - 3 x^{3} y

f (x) = x^{2} - 1, - 4 \leq x \leq 4

\frac{d}{d x} (6 x^{2} y - 2)

What is the extreme f(x)=x^4-2x^3,-2<= x<= 2 ?
The extreme f(x)=x^4-2x^3,-2<= x<= 2 is Maximum(-2,32),Saddle(0,0),Minimum(3/2 ,-27/16),Maximum(2,0)