What is the extreme f(x)=(x^3)/3-x^2-8x-4 ?

The extreme f(x)=(x^3)/3-x^2-8x-4 is Maximum(-2, 16/3),Minimum(4,-92/3)

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extreme f(x)=(x^3)/3-x^2-8x-4

extreme

Maximum (- 2, \frac{16}{3}), Minimum (4, - \frac{92}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} - x^{2} - 8 x - 4 : \frac{16}{3}

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

Plug

x = 4

into

\frac{x ^{3}}{3} - x^{2} - 8 x - 4 : - \frac{92}{3}

Minimum (4, - \frac{92}{3})

Maximum (- 2, \frac{16}{3}), Minimum (4, - \frac{92}{3})

What is the extreme f(x)=(x^3)/3-x^2-8x-4 ?
The extreme f(x)=(x^3)/3-x^2-8x-4 is Maximum(-2, 16/3),Minimum(4,-92/3)