What is the extreme f(x)=2x^4+8x^3+5x^2 ?

The extreme f(x)=2x^4+8x^3+5x^2 is Minimum(-5/2 ,-125/8),Maximum(-1/2 , 3/8),Minimum(0,0)

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extreme f(x)=2x^4+8x^3+5x^2

extreme points

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

Minimum (- \frac{5}{2}, - \frac{125}{8}), Maximum (- \frac{1}{2}, \frac{3}{8}), Minimum (0, 0)

Solution steps

f^{'} (x) = 8 x^{3} + 24 x^{2} + 10 x

Find intervals:

Decreasing

: - \infty < x < - \frac{5}{2},

Increasing

: - \frac{5}{2} < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{5}{2}

into

2 x^{4} + 8 x^{3} + 5 x^{2} : - \frac{125}{8}

Minimum (- \frac{5}{2}, - \frac{125}{8})

Plug

x = - \frac{1}{2}

into

2 x^{4} + 8 x^{3} + 5 x^{2} : \frac{3}{8}

Maximum (- \frac{1}{2}, \frac{3}{8})

Plug

x = 0

into

2 x^{4} + 8 x^{3} + 5 x^{2} : 0

Minimum (0, 0)

Minimum (- \frac{5}{2}, - \frac{125}{8}), Maximum (- \frac{1}{2}, \frac{3}{8}), Minimum (0, 0)

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

\frac{3}{7 + x}

\frac{x + 5}{x - 2}

f (x) = - 0.5 x + 6 + 2

What is the extreme f(x)=2x^4+8x^3+5x^2 ?
The extreme f(x)=2x^4+8x^3+5x^2 is Minimum(-5/2 ,-125/8),Maximum(-1/2 , 3/8),Minimum(0,0)