Answer
$$
f(x) =\frac{2^{x}}{x}
$$
A relative minimum occurs at $x=\frac{1}{\ln 2 } \approx 1.44 $ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(1.44 ) &=\frac{2^{(1.44 )}}{(1.44 )} \\
&=1.884 \\
\end{aligned}
$$
Work Step by Step
$$
f(x) =\frac{2^{x}}{x}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=\frac{(x) \ln 2\left(2^{x}\right)-2^{x}(1)}{x^{2}} \\
&=\frac{2^{x}(x \ln 2-1)}{x^{2}}
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =\frac{2^{x}(x \ln 2-1)}{x^{2}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x \ln 2-1&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x \ln 2 & =1 \\
x & =\frac{1}{\ln 2 } \approx 1.44 \\
\end{aligned}
$$
Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $x^{2}$ equal to 0, when $x=0$. But $f(x) $ is not defined also at $x=0 $. So,
The only critical number is : $x =\frac{1}{\ln 2 } \approx 1.44 $.
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
$$
(-\infty, 1.44 ), \quad\quad (1.44 , \infty ) .
$$
(1)
Test a number in the interval $(-\infty, 1.44 )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=\frac{2^{(1)}((1) \ln 2-1)}{(1)^{2}}\\
&=2\left(\ln \left(2\right)-1\right) \\
&\approx -0.6137\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty , 1.44 )$.
(2)
Test a number in the interval $(1.44 , \infty )$ say $2$:
$$
\begin{aligned}
f^{\prime}(2) &=\frac{2^{(2)}((2) \ln 2-1)}{(2)^{2}}\\
&=2\ln \left(2\right)-1 \\
&\approx 0.386\\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(1.44 , \infty )$.
From (1), (2) we find that:
The function $f$ is decreasing on interval $ ( -\infty ,1.44) , $ and it is increasing on interval $ (1.44 ,\infty ) . $ So, a relative minimum occurs at $x=1.44 $ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(1.44 ) &=\frac{2^{(1.44 )}}{(1.44 )} \\
&=1.884 \\
\end{aligned}
$$
