What is the extreme f(x)=4sqrt(x^2+1)-x ?

The extreme f(x)=4sqrt(x^2+1)-x is Minimum(1/(sqrt(15)),sqrt(15))

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extreme f(x)=4sqrt(x^2+1)-x

extreme points

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

Minimum (\frac{1}{15}, 15)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{1}{15},

Increasing

: \frac{1}{15} < x < \infty

Plug

x = \frac{1}{15}

into

4 x^{2} + 1 - x : 15

Minimum (\frac{1}{15}, 15)

x^{2} - 4 x + 10

3 x + 5 y = 0

f (x) = (x - 5)^{2} + 2

f (x) = \frac{5 - x}{x ^{2} - 9}

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

What is the extreme f(x)=4sqrt(x^2+1)-x ?
The extreme f(x)=4sqrt(x^2+1)-x is Minimum(1/(sqrt(15)),sqrt(15))