What is the extreme f(x)= 1/3*x^3-1/2 x^2-6x+5 ?

The extreme f(x)= 1/3*x^3-1/2 x^2-6x+5 is Maximum(-2, 37/3),Minimum(3,-17/2)

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extreme f(x)= 1/3*x^3-1/2 x^2-6x+5

extreme points

f (x) = \frac{1}{3} \cdot x^{3} - \frac{1}{2} x^{2} - 6 x + 5

Maximum (- 2, \frac{37}{3}), Minimum (3, - \frac{17}{2})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 2

into

\frac{1}{3} \cdot x^{3} - \frac{1}{2} x^{2} - 6 x + 5 : \frac{37}{3}

Maximum (- 2, \frac{37}{3})

Plug

x = 3

into

\frac{1}{3} \cdot x^{3} - \frac{1}{2} x^{2} - 6 x + 5 : - \frac{17}{2}

Minimum (3, - \frac{17}{2})

Maximum (- 2, \frac{37}{3}), Minimum (3, - \frac{17}{2})

\frac{e ^{2 \cdot x}}{6 + 5 \cdot e ^{2 \cdot x}}

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

f (x) = - 2 x^{2} - 10 x + 5

\frac{112}{( x + 5 ) ^{3}}

f (x) = \frac{1}{5} x^{5} - \frac{1}{2} x^{4}

What is the extreme f(x)= 1/3*x^3-1/2 x^2-6x+5 ?
The extreme f(x)= 1/3*x^3-1/2 x^2-6x+5 is Maximum(-2, 37/3),Minimum(3,-17/2)