What is the extreme f(x)=x^3+5x^2-4 ?

The extreme f(x)=x^3+5x^2-4 is Maximum(-10/3 , 392/27),Minimum(0,-4)

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extreme f(x)=x^3+5x^2-4

extreme points

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

Maximum (- \frac{10}{3}, \frac{392}{27}), Minimum (0, - 4)

Solution steps

f^{'} (x) = 3 x^{2} + 10 x

Find intervals:

Increasing

: - \infty < x < - \frac{10}{3},

Decreasing

: - \frac{10}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{10}{3}

into

x^{3} + 5 x^{2} - 4 : \frac{392}{27}

Maximum (- \frac{10}{3}, \frac{392}{27})

Plug

x = 0

into

x^{3} + 5 x^{2} - 4 : - 4

Minimum (0, - 4)

Maximum (- \frac{10}{3}, \frac{392}{27}), Minimum (0, - 4)

f (x) = \frac{1}{x ^{2} - x - 56}

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

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

ln (x - 5)

x^{2} y^{2} + x y = 1

What is the extreme f(x)=x^3+5x^2-4 ?
The extreme f(x)=x^3+5x^2-4 is Maximum(-10/3 , 392/27),Minimum(0,-4)