What is the extreme f(x)= 1/7 x^7-x ?

The extreme f(x)= 1/7 x^7-x is Maximum(-1, 6/7),Minimum(1,-6/7)

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extreme f(x)= 1/7 x^7-x

extreme points

f (x) = \frac{1}{7} x^{7} - x

Maximum (- 1, \frac{6}{7}), Minimum (1, - \frac{6}{7})

Solution steps

f^{'} (x) = x^{6} - 1

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{1}{7} x^{7} - x : \frac{6}{7}

Maximum (- 1, \frac{6}{7})

Plug

x = 1

into

\frac{1}{7} x^{7} - x : - \frac{6}{7}

Minimum (1, - \frac{6}{7})

Maximum (- 1, \frac{6}{7}), Minimum (1, - \frac{6}{7})

f (x, y) = \frac{1}{x} + x y - \frac{8}{y}

f (x) = e^{x y} - x

y = 3 x^{2} - 12 x + 13

f (x) = \frac{1}{4} x^{4} - 2 x^{3} + 3

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

What is the extreme f(x)= 1/7 x^7-x ?
The extreme f(x)= 1/7 x^7-x is Maximum(-1, 6/7),Minimum(1,-6/7)