What is the extreme f(x)=(x^3)/3+(x^2)/2-6x+10 ?

The extreme f(x)=(x^3)/3+(x^2)/2-6x+10 is Maximum(-3, 47/2),Minimum(2, 8/3)

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extreme f(x)=(x^3)/3+(x^2)/2-6x+10

extreme

Maximum (- 3, \frac{47}{2}), Minimum (2, \frac{8}{3})

Solution steps

f^{'} (x) = x^{2} + x - 6

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 3

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 10 : \frac{47}{2}

Maximum (- 3, \frac{47}{2})

Plug

x = 2

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 10 : \frac{8}{3}

Minimum (2, \frac{8}{3})

Maximum (- 3, \frac{47}{2}), Minimum (2, \frac{8}{3})

What is the extreme f(x)=(x^3)/3+(x^2)/2-6x+10 ?
The extreme f(x)=(x^3)/3+(x^2)/2-6x+10 is Maximum(-3, 47/2),Minimum(2, 8/3)