What is the extreme f(x)=6sqrt(x^2+1)-x ?

The extreme f(x)=6sqrt(x^2+1)-x is Minimum(1/(sqrt(35)),sqrt(35))

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extreme f(x)=6sqrt(x^2+1)-x

extreme points

f (x) = 6 x^{2} + 1 - x

Minimum (\frac{1}{35}, 35)

Solution steps

f^{'} (x) = \frac{6 x}{x ^{2} + 1} - 1

Find intervals:

Decreasing

: - \infty < x < \frac{1}{35},

Increasing

: \frac{1}{35} < x < \infty

Plug

x = \frac{1}{35}

into

6 x^{2} + 1 - x : 35

Minimum (\frac{1}{35}, 35)

What is the extreme f(x)=6sqrt(x^2+1)-x ?
The extreme f(x)=6sqrt(x^2+1)-x is Minimum(1/(sqrt(35)),sqrt(35))