Single Variable Calculus, Chapter 7, 7.3-1, Section 7.3-1, Problem 62

At what interval is the curve $y = 2e^x - e^{-3x}$ concave downward?


$
\begin{equation}
\begin{aligned}

\text{If } y =& 2e^x - e^{-3x}, \text{ then}
\\
\\
y' =& 2e^x - e^{-3x} (-3) = 2e^x + 3e^{-3x}
\\
\\
\text{then,} &
\\
\\
y" =& 2e^x + 3e^{-3x} (-3)
\\
\\
y" =& 2e^x - 9e^{-3x}


\end{aligned}
\end{equation}
$


To know the intervals of concavity, we set $y" = 0$ and get the inflection points.


$
\begin{equation}
\begin{aligned}

& y" = 0 = 2e^x - 9e^{-3x}
\\
\\
& 0 = 2e^x - 9e^{-3x}
\\
\\
& 9e^{-3x} = 2e^x
\\
\\
& e^{(x - (-3x))} = \frac{9}{2}
\\
\\
& e^{4x} = \frac{9}{2}

\end{aligned}
\end{equation}
$


If we take the natural logarithm of both sides


$
\begin{equation}
\begin{aligned}

lne^{4x} =& lm \frac{9}{2}
\\
\\
4x =& ln \frac{9}{2}

\end{aligned}
\end{equation}
$


The inflection point is at..

$\displaystyle x = \frac{ln(4.5)}{4}$

The interval of concavity is..


$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
\hline\\
\displaystyle x < \frac{ln (4.5)}{4} & - & \text{Downward} \\
\hline\\
\displaystyle x > \frac{ln (4.5)}{4} & + & \text{Upward}\\
\hline
\end{array}
$


Therefore, the curve $y = 2e^x - e^{-3x}$ has downward concavity at interval $\displaystyle \left( - \infty, \frac{ln (4.5)}{4} \right)$