What is the extreme f(x)=8sqrt(x^2+1)-x ?

The extreme f(x)=8sqrt(x^2+1)-x is Minimum(1/(3sqrt(7)),3sqrt(7))

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extreme f(x)=8sqrt(x^2+1)-x

extreme points

f (x) = 8 x^{2} + 1 - x

Minimum (\frac{1}{3 7}, 37)

Solution steps

f^{'} (x) = \frac{8 x}{x ^{2} + 1} - 1

Find intervals:

Decreasing

: - \infty < x < \frac{1}{3 7},

Increasing

: \frac{1}{3 7} < x < \infty

Plug

x = \frac{1}{3 7}

into

8 x^{2} + 1 - x : 37

Minimum (\frac{1}{3 7}, 37)

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

y = x^{2} - 2 x - 80

f (x) = \frac{x - 4}{x ^{2} - 9}

f (x) = - \frac{1}{3} sin (3 x)

(2, - 11), (- 9, 0)

What is the extreme f(x)=8sqrt(x^2+1)-x ?
The extreme f(x)=8sqrt(x^2+1)-x is Minimum(1/(3sqrt(7)),3sqrt(7))