Answer
$$
f(x) =\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}
$$
*)
The function is increasing on intervals $ (-\infty ,0 ), $
**)
The function is decreasing on interval $(0 , \infty ) .$
Work Step by Step
The standard normal probability function is defined by:
$$
f(x) =\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}(-x) \\
&=\frac{-x}{\sqrt{2 \pi}} e^{-x^{2} / 2}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime} =0 $. Now, solve the equation $f^{\prime} =0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =\frac{-x}{\sqrt{2 \pi}} e^{-x^{2} / 2} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=0
\end{aligned}
$$
So, the only critical number is $x=0$, and this number divides the number line into two intervals:
$$
(-\infty, 0 ), \quad (0 , \infty ) .
$$
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. Now, check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, 0 ), \quad (0 , \infty ) .
$$
(1)
Test a number in the interval $(-\infty, 0) $ say $-1 $:
$$
\begin{aligned}
f^{\prime}(-1) &=\frac{-(-1)}{\sqrt{2 \pi}} e^{-(-1)^{2} / 2}\\
&=\frac{1}{\sqrt{2}\sqrt{\pi }\sqrt{e}} \\
&\approx 0.24197 \\
& \gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, 0)$.
(2)
Test a number in the interval $(0 , \infty ) $ say $ 1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{-(1)}{\sqrt{2 \pi}} e^{-(1)^{2} / 2}\\
&=-\frac{1}{\sqrt{2}\sqrt{\pi }\sqrt{e}} \\
&\approx -0.24197 \\
& \lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is decreasing on $(0 , \infty ) $.
So,
*)
The function is increasing on intervals $ (-\infty ,0 ), $
**)
The function is decreasing on interval $(0 , \infty ) .$
