Answer
$$
y=\sqrt {x^{2}+1}
$$
(a)
The critical number : $x=0$
(b)
The function is increasing , on interval $(0, \infty)$
(c)
The function is decreasing on interval $(-\infty, 0)$ .
Work Step by Step
$$
y=\sqrt {x^{2}+1}
$$
We find $y^{\prime} $ first, using the power rule and the chain rule.
$$
\begin{aligned}
y^{\prime} &=\frac{1}{2}\left(x^{2}+1\right)^{-1 / 2}(2 x) \\
&=x\left(x^{2}+1\right)^{-1 / 2} \\
&=\frac{x}{\sqrt{x^{2}+1}}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $y^{\prime} =0 $.
Now, solve the equation $y^{\prime} =0 $ to get
$$
\begin{aligned}
y^{\prime} =\frac{x}{\sqrt{x^{2}+1}}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =0
\end{aligned}
$$
Next, we find any values of $x$ where fails to exist. This
occurs whenever the denominator of $y^{\prime}$ is $0$, so set the denominator equal to $0$. Since there are no values of $x$ where fails to exist, the only critical number is $x=0$ .
We observe, the number $0$ divides the number line into two intervals: $(-\infty, 0)$ and $(0, \infty)$. use a test point in each of these intervals to find that sign of $y^{\prime} $ (This can also be determined by observing that $y^{\prime}(x) $ is the quotient of $x$, which is negative or positive , and $\sqrt{x^{2}+1}$ , which is always positive.) This means that the function $f $ is decreasing on $(-\infty, 0)$ and is increasing on$(0, \infty)$.
So,
(a)
The critical number : $x=0$
(b)
The function is increasing , on interval $(0, \infty)$
(c)
The function is decreasing on interval $(-\infty, 0)$ .
