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

The extreme f(x)=(x^3)/3+x^2-8x+2 is Maximum(-4, 86/3),Minimum(2,-22/3)

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

extreme

Maximum (- 4, \frac{86}{3}), Minimum (2, - \frac{22}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 4

into

\frac{x ^{3}}{3} + x^{2} - 8 x + 2 : \frac{86}{3}

Maximum (- 4, \frac{86}{3})

Plug

x = 2

into

\frac{x ^{3}}{3} + x^{2} - 8 x + 2 : - \frac{22}{3}

Minimum (2, - \frac{22}{3})

Maximum (- 4, \frac{86}{3}), Minimum (2, - \frac{22}{3})

What is the extreme f(x)=(x^3)/3+x^2-8x+2 ?
The extreme f(x)=(x^3)/3+x^2-8x+2 is Maximum(-4, 86/3),Minimum(2,-22/3)