What is the extreme f(x)= 1/3 x^3+1/2 x^2-6x+8 ?

The extreme f(x)= 1/3 x^3+1/2 x^2-6x+8 is Maximum(-3, 43/2),Minimum(2, 2/3)

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extreme f(x)= 1/3 x^3+1/2 x^2-6x+8

extreme points

f (x) = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8

Maximum (- 3, \frac{43}{2}), Minimum (2, \frac{2}{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{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8 : \frac{43}{2}

Maximum (- 3, \frac{43}{2})

Plug

x = 2

into

\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8 : \frac{2}{3}

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

Maximum (- 3, \frac{43}{2}), Minimum (2, \frac{2}{3})

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

f (x, y) = 9 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = x^{2} e^{- x}, - 1 \leq x \leq 1

f (x) = x^{4} + 4 x^{3} - 2

f (x, y) = 36 - 9 x^{2} - 4 y^{2}

What is the extreme f(x)= 1/3 x^3+1/2 x^2-6x+8 ?
The extreme f(x)= 1/3 x^3+1/2 x^2-6x+8 is Maximum(-3, 43/2),Minimum(2, 2/3)