What is the extreme f(x)=16x^3(x+1)^2 ?

The extreme f(x)=16x^3(x+1)^2 is Maximum(-1,0),Minimum(-3/5 ,-1728/3125),Saddle(0,0)

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extreme f(x)=16x^3(x+1)^2

extreme points

f (x) = 16 x^{3} (x + 1)^{2}

Maximum (- 1, 0), Minimum (- \frac{3}{5}, - \frac{1728}{3125}), Saddle (0, 0)

Solution steps

f^{'} (x) = 16 (5 x^{4} + 8 x^{3} + 3 x^{2})

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

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

Increasing

: - \frac{3}{5} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

16 x^{3} (x + 1)^{2} : 0

Maximum (- 1, 0)

Plug

x = - \frac{3}{5}

into

16 x^{3} (x + 1)^{2} : - \frac{1728}{3125}

Minimum (- \frac{3}{5}, - \frac{1728}{3125})

Plug

x = 0

into

16 x^{3} (x + 1)^{2} : 0

Saddle (0, 0)

Maximum (- 1, 0), Minimum (- \frac{3}{5}, - \frac{1728}{3125}), Saddle (0, 0)

2 x^{\frac{3}{2}} - 9 x + 6

f (x) = 5 x^{2} - 8

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

f (x = - 1, y = 1) = x^{2} + y^{2} + 2 x - 2 y + 2

f (x) = 7 x + 2 y - 8

What is the extreme f(x)=16x^3(x+1)^2 ?
The extreme f(x)=16x^3(x+1)^2 is Maximum(-1,0),Minimum(-3/5 ,-1728/3125),Saddle(0,0)