Answer
$$
f(x) =x+8^{-x}
$$
A relative minimum occurs at $x=\frac{\ln (\ln 8)}{\ln 8} \approx 0.35 $ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(0.35) &=(0.35)+8^{-(0.35)} \\
&=0.83 \\
\end{aligned}
$$
Work Step by Step
$$
f(x) =x+8^{-x}
$$
To find the intervals where the function is increasing and decreasing.
First, we find $f^{\prime}(x) $,
$$
\begin{aligned}
f^{\prime}(x) &=1+(\ln 8) 8^{-x}(-1) \\
&=1-\frac{\ln 8}{8^{x}}
\end{aligned}
$$
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
$$
\begin{aligned}
f^{\prime}(x) =1-\frac{\ln 8}{8^{x}} &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
\frac{\ln 8}{8^{x}}&=1\\
\Rightarrow\quad\quad\quad\quad\quad\\
\ln 8 & =8^{x} \\
x \ln 8 &=\ln (\ln 8) \\
x &=\frac{\ln (\ln 8)}{\ln 8} \approx 0.35
\end{aligned}
$$
The only critical number is : $x =\frac{\ln (\ln 8)}{\ln 8} \approx 0.35.$
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
$$
(-\infty, 0.35 ), \quad\quad (0.35 , \infty ) .
$$
(1)
Test a number in the interval $(-\infty, 0.35 )$ say $0$:
$$
\begin{aligned}
f^{\prime}(0) &=1-\frac{\ln 8}{8^{(0)}}\\
&=1-3\ln \left(2\right)\\
&\approx -1.0794\\
&\lt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty , 0.35 )$.
(2)
Test a number in the interval $(0.35 , \infty )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) &=1-\frac{\ln 8}{8^{(1)}}\\
&=1-\frac{3\ln \left(2\right)}{8}\\
&\approx 0.740\\
&\gt 0
\end{aligned}
$$
we see that $f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(0.35 , \infty )$.
From (1), (2) we find that:
(*)
The function $f$ is decreasing on interval $ ( -\infty ,0.35) , $ and it is increasing on interval $ (0.35 ,\infty ) . $ So, a relative minimum occurs at $x=0.35 $ and the value of the function $f$ at this value is given by:
$$
\begin{aligned}
f(0.35) &=(0.35)+8^{-(0.35)} \\
&=0.83 \\
\end{aligned}
$$
