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

The extreme f(x)= 1/3 x^3-x^2-4 is Maximum(0,-4),Minimum(2,-16/3)

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

extreme points

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

Maximum (0, - 4), Minimum (2, - \frac{16}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, - 4)

Plug

x = 2

into

\frac{1}{3} x^{3} - x^{2} - 4 : - \frac{16}{3}

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

Maximum (0, - 4), Minimum (2, - \frac{16}{3})

7 cos (10 x - 6 π) + 8

\frac{x}{x + 7}

s^{3}

\frac{x}{x + 1}

\frac{7 x - 4 y}{4} = x + 2

What is the extreme f(x)= 1/3 x^3-x^2-4 ?
The extreme f(x)= 1/3 x^3-x^2-4 is Maximum(0,-4),Minimum(2,-16/3)