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

The extreme f(x)=x^3+x^2-5 is Maximum(-2/3 , 4/27-5),Minimum(0,-5)

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extreme f(x)=x^3+x^2-5

extreme points

f (x) = x^{3} + x^{2} - 5

Maximum (- \frac{2}{3}, \frac{4}{27} - 5), Minimum (0, - 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} + x^{2} - 5 : \frac{4}{27} - 5

Maximum (- \frac{2}{3}, \frac{4}{27} - 5)

Plug

x = 0

into

x^{3} + x^{2} - 5 : - 5

Minimum (0, - 5)

Maximum (- \frac{2}{3}, \frac{4}{27} - 5), Minimum (0, - 5)

f (x, y) = 25 - x^{2} - y^{2}

- 7 (x + 3)^{2} (x - 1) (x - 5)

f (x) = 2 x^{2} - 5 x y + 3 y^{4} + 5

f (x, y) = \frac{1}{x - y}

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

What is the extreme f(x)=x^3+x^2-5 ?
The extreme f(x)=x^3+x^2-5 is Maximum(-2/3 , 4/27-5),Minimum(0,-5)