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

The extreme f(x)=2x^4+8x^3 is Minimum(-3,-54),Saddle(0,0)

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

extreme points

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

Minimum (- 3, - 54), Saddle (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

2 x^{4} + 8 x^{3} : - 54

Minimum (- 3, - 54)

Plug

x = 0

into

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

Saddle (0, 0)

Minimum (- 3, - 54), Saddle (0, 0)

in v erse f (x) = \frac{7}{x + 4}

in v erse f (x) = \frac{x + 8}{x - 2}

d o main ln (x) + ln (4 - x)

in v erse f (x) = x + 5 - 1

d o main x^{6} - \frac{6}{5} x^{5}

What is the extreme f(x)=2x^4+8x^3 ?
The extreme f(x)=2x^4+8x^3 is Minimum(-3,-54),Saddle(0,0)