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 points

f (x) = (\frac{1}{12} x^{3} + x^{2} + 3 x + \frac{26}{3})

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)

f (x) = - 5 x^{3} + 45 x

f (x) = - 0.020724 x^{2} + 0.3692 x - 0.6287

f (x, y) = x^{3} - y^{2} - 3 x + 12 y

f (x) = (x^{2} - 12) \cdot e^{- 2 x}

f (x) = 49 x^{2} - 10 x + 17

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)