What is the extreme 5x^2(x+1)^{1/2} ?

The extreme 5x^2(x+1)^{1/2} is Minimum(-1,0),Maximum(-4/5 ,(16)/(5*5^{1/2)}),Minimum(0,0)

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extreme 5x^2(x+1)^{1/2}

extreme points

5 x^{2} (x + 1)^{\frac{1}{2}}

Minimum (- 1, 0), Maximum (- \frac{4}{5}, \frac{16}{5 \cdot 5 ^{\frac{1}{2}}}), Minimum (0, 0)

Solution steps

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

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} (x + 1)^{\frac{1}{2}} : 0

Minimum (- 1, 0)

Plug

x = - \frac{4}{5}

into

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

Maximum (- \frac{4}{5}, \frac{16}{5 \cdot 5 ^{\frac{1}{2}}})

Plug

x = 0

into

5 x^{2} (x + 1)^{\frac{1}{2}} : 0

Minimum (0, 0)

Minimum (- 1, 0), Maximum (- \frac{4}{5}, \frac{16}{5 \cdot 5 ^{\frac{1}{2}}}), Minimum (0, 0)

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

f (x) = x^{3} - 8 y^{3} + 4 x y - 8

7 + 12 x - x^{3}

f (x) = x + e^{- 3 x}, - 3 \leq x \leq 4

f (x) = x e^{- e}, 0 \leq x \leq 2

What is the extreme 5x^2(x+1)^{1/2} ?
The extreme 5x^2(x+1)^{1/2} is Minimum(-1,0),Maximum(-4/5 ,(16)/(5*5^{1/2)}),Minimum(0,0)