# `f(x) = sqrt(x^2 + 1) - x` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the...

`f(x) = sqrt(x^2 + 1) - x` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of `f`.

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### 2 Answers

... and f hasn't inflection points.

The function is defined everywhere and is infinitely differentiable.

(a) so no vertical asymptotes

At right asymptote y=0:

`lim_(x->+oo)f(x) = lim_(x->+oo)(sqrt(x^2+1)-x)` =

`lim_(x->+oo)((x^2+1-x^2)/(sqrt(x^2+1)+x)) =lim_(x->+oo)(1/(sqrt(x^2+1)+x)) = 0. `

When x->-oo, f(x)->+oo, no horizontal asymptote at left (slanted is).

(b, c) `f'(x)=1/(2*sqrt(x^2+1))*2x-1 = x/sqrt(x^2+1) - 1`

which is <0, so f decreases on `(-oo, +oo)` and has no minimums and maximums.

d) `f''(x)=(1* sqrt(x^2+1) - x*(1/(2*sqrt(x^2+1))*2x))/(x^2+1)=( x^2+1- x^2)/( sqrt(x^2+1)* (x^2+1))=1/( sqrt(x^2+1)* (x^2+1)).`

which is >0 on `(-oo, +oo)` so f is concave upward on `(-oo, +oo)` .

e)