What is the extreme f(x)= 1/4 x^4-x^3-2x^2 ?

The extreme f(x)= 1/4 x^4-x^3-2x^2 is Minimum(-1,-3/4),Maximum(0,0),Minimum(4,-32)

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extreme f(x)= 1/4 x^4-x^3-2x^2

extreme points

f (x) = \frac{1}{4} x^{4} - x^{3} - 2 x^{2}

Minimum (- 1, - \frac{3}{4}), Maximum (0, 0), Minimum (4, - 32)

Solution steps

f^{'} (x) = x^{3} - 3 x^{2} - 4 x

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, - \frac{3}{4})

Plug

x = 0

into

\frac{1}{4} x^{4} - x^{3} - 2 x^{2} : 0

Maximum (0, 0)

Plug

x = 4

into

\frac{1}{4} x^{4} - x^{3} - 2 x^{2} : - 32

Minimum (4, - 32)

Minimum (- 1, - \frac{3}{4}), Maximum (0, 0), Minimum (4, - 32)

f (x) = 3 x - 2 y

f (x) = - \frac{12}{x ^{4}} + \frac{14}{x ^{3}}

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1, - 6 \leq x \leq 4

f (x) = x^{2} + \frac{x ^{3}}{3} - (\frac{x}{4})^{4}

f (x) = x e^{- 7 x}, 0 \leq x \leq 2

What is the extreme f(x)= 1/4 x^4-x^3-2x^2 ?
The extreme f(x)= 1/4 x^4-x^3-2x^2 is Minimum(-1,-3/4),Maximum(0,0),Minimum(4,-32)