Answer
(a) $y = 0$ is a horizontal asymptote.
(b) $f$ is decreasing on the interval $(-\infty,\infty)$
(c) There is no local maximum or local minimum.
(d) The graph is concave up on this interval: $(-\infty, \infty)$
There are no points of inflection.
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = \sqrt{x^2+1}-x$
$f(x)$ is defined for all $x$ so there are no vertical asymptotes.
$\lim\limits_{x \to -\infty}f(x) = \infty$
$\lim\limits_{x \to \infty}f(x) = 0$
$y = 0$ is a horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = \frac{x}{\sqrt{x^2+1}}-1 = 0$
$\frac{x}{\sqrt{x^2+1}} = 1$
There are no solutions to this equation since $\sqrt{x^2+1} \gt x$ for all values of $x$
When $x \lt 0~~$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(-\infty,\infty)$
When $x \gt 0$ then $f'(x) \gt 0$
$f$ is not increasing on any interval.
(c) There are no values $x$ such that $f'(x) = 0$
There is no local maximum or local minimum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{\sqrt{x^2+1}-(x)(\frac{x}{\sqrt{x^2+1}})}{x^2+1}$
$f''(x) = \frac{\frac{x^2+1}{\sqrt{x^2+1}}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}$
$f''(x) = \frac{1}{(x^2+1)^{3/2}}$
There are no values of $x$ such that $f''(x) = 0$ since $f''(x) \gt 0$ for all values of $x$
The graph is concave down when $f''(x) \lt 0$
The graph is not concave down on any interval.
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-\infty, \infty)$
There are no points of inflection.
(e) We can see a sketch of the graph below.
