What is the extreme f(x)=5x^2sqrt(1+x) ?

The extreme f(x)=5x^2sqrt(1+x) is Minimum(-1,0),Maximum(-4/5 ,(16)/(5sqrt(5))),Minimum(0,0)

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extreme f(x)=5x^2sqrt(1+x)

extreme points

f (x) = 5 x^{2} 1 + x

Minimum (- 1, 0), Maximum (- \frac{4}{5}, \frac{16}{5 5}), Minimum (0, 0)

Solution steps

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

Find intervals:

Increasing

: - 1 < x < - \frac{4}{5},

Decreasing

: - \frac{4}{5} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

5 x^{2} 1 + x : 0

Minimum (- 1, 0)

Plug

x = - \frac{4}{5}

into

5 x^{2} 1 + x : \frac{16}{5 5}

Maximum (- \frac{4}{5}, \frac{16}{5 5})

Plug

x = 0

into

5 x^{2} 1 + x : 0

Minimum (0, 0)

Minimum (- 1, 0), Maximum (- \frac{4}{5}, \frac{16}{5 5}), Minimum (0, 0)

f (x) = x^{3} - \frac{3}{2} x^{2} - 18 x

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

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

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

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

What is the extreme f(x)=5x^2sqrt(1+x) ?
The extreme f(x)=5x^2sqrt(1+x) is Minimum(-1,0),Maximum(-4/5 ,(16)/(5sqrt(5))),Minimum(0,0)