What is the extreme f(x)=(x^4)/4-(8x^3)/3-32x^2+512x ?

The extreme f(x)=(x^4)/4-(8x^3)/3-32x^2+512x is Minimum(-8,-11264/3),Saddle(8, 5120/3)

Popular >

extreme f(x)=(x^4)/4-(8x^3)/3-32x^2+512x

extreme points

f (x) = \frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} - 32 x^{2} + 512 x

Minimum (- 8, - \frac{11264}{3}), Saddle (8, \frac{5120}{3})

Solution steps

f^{'} (x) = x^{3} - 8 x^{2} - 64 x + 512

Find intervals:

Decreasing

: - \infty < x < - 8,

Increasing

: - 8 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = - 8

into

\frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} - 32 x^{2} + 512 x : - \frac{11264}{3}

Minimum (- 8, - \frac{11264}{3})

Plug

x = 8

into

\frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} - 32 x^{2} + 512 x : \frac{5120}{3}

Saddle (8, \frac{5120}{3})

Minimum (- 8, - \frac{11264}{3}), Saddle (8, \frac{5120}{3})

e^{3 x} (4 - x)

f (x) = 5 x^{2} + 5 y^{2} + 20 x - 10 y + 40

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

\frac{ln ( x )}{x ^{8}}

f (x) = (sin (x)) (cos (x))

What is the extreme f(x)=(x^4)/4-(8x^3)/3-32x^2+512x ?
The extreme f(x)=(x^4)/4-(8x^3)/3-32x^2+512x is Minimum(-8,-11264/3),Saddle(8, 5120/3)