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

The extreme f(x)=x^3+x^2-5x-5 is Maximum(-5/3 , 40/27),Minimum(1,-8)

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

extreme points

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

Maximum (- \frac{5}{3}, \frac{40}{27}), Minimum (1, - 8)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{5}{3},

Decreasing

: - \frac{5}{3} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - \frac{5}{3}

into

x^{3} + x^{2} - 5 x - 5 : \frac{40}{27}

Maximum (- \frac{5}{3}, \frac{40}{27})

Plug

x = 1

into

x^{3} + x^{2} - 5 x - 5 : - 8

Minimum (1, - 8)

Maximum (- \frac{5}{3}, \frac{40}{27}), Minimum (1, - 8)

f (x, y) = 5 x^{2} + 3 x y + e^{x y}

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

f (x) = 0.01 x^{3} + 0.45 x^{2} + 2.43 x + 300

f (x) = \frac{1}{9} \cdot (\frac{x ^{3}}{x + 2})

f (x) = x^{3} - 5 x^{2} + 3 x + 10, - 2 \leq x \leq 4

What is the extreme f(x)=x^3+x^2-5x-5 ?
The extreme f(x)=x^3+x^2-5x-5 is Maximum(-5/3 , 40/27),Minimum(1,-8)