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

The extreme f(x)= 1/3 x^3+x^2-3x is Maximum(-3,9),Minimum(1,-5/3)

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

extreme points

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

Maximum (- 3, 9), Minimum (1, - \frac{5}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 3

into

\frac{1}{3} x^{3} + x^{2} - 3 x : 9

Maximum (- 3, 9)

Plug

x = 1

into

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

Minimum (1, - \frac{5}{3})

Maximum (- 3, 9), Minimum (1, - \frac{5}{3})

\frac{2 x + 3}{x ( x ^{2} + 2 x - 3 )}

f (x) = \frac{3 x + 1}{( x - 1 )}

f (x) = 5 \cdot cos (\frac{2 \cdot π \cdot x}{3})

y = - tan (x - \frac{3 π}{2})

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

What is the extreme f(x)= 1/3 x^3+x^2-3x ?
The extreme f(x)= 1/3 x^3+x^2-3x is Maximum(-3,9),Minimum(1,-5/3)