What is the extreme f(x)=(1/12 x^3+x^2+3x+26/3) ?

The extreme f(x)=(1/12 x^3+x^2+3x+26/3) is Maximum(-6, 26/3),Minimum(-2,6)

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extreme f(x)=(1/12 x^3+x^2+3x+26/3)

extreme

Maximum (- 6, \frac{26}{3}), Minimum (- 2, 6)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 6

into

\frac{1}{12} x^{3} + x^{2} + 3 x + \frac{26}{3} : \frac{26}{3}

Maximum (- 6, \frac{26}{3})

Plug

x = - 2

into

\frac{1}{12} x^{3} + x^{2} + 3 x + \frac{26}{3} : 6

Minimum (- 2, 6)

Maximum (- 6, \frac{26}{3}), Minimum (- 2, 6)

What is the extreme f(x)=(1/12 x^3+x^2+3x+26/3) ?
The extreme f(x)=(1/12 x^3+x^2+3x+26/3) is Maximum(-6, 26/3),Minimum(-2,6)