Given
[tex]f(x)=xe^{7x}[/tex]
Calculate the second derivative of f(x), as shown below
[tex]\begin{gathered} \Rightarrow f^{\prime}(x)=e^{7x}+7xe^{7x} \\ and \\ \Rightarrow f^{\prime}^{\prime}(x)=7e^{7x}+7(e^{7x}+7xe^{7x}) \\ \Rightarrow f^{\prime}^{\prime}(x)=14e^{7x}+49xe^{7x} \end{gathered}[/tex]
Then, find the interval such that f''(x)>0 in order to find where f(x) is concave up,
[tex]\begin{gathered} 14e^{7x}+49xe^{7x}>0 \\ \Rightarrow2e^{7x}+7x*e^{7x}>0 \\ and \\ e{}^{7x}>0,x\in\Re \end{gathered}[/tex]
Then,
[tex]\begin{gathered} 2e^{7x}>-7xe^{7x} \\ \Rightarrow2>-7x \\ \Rightarrow x>-\frac{2}{7} \end{gathered}[/tex]
Therefore, f(x) is concave up when x in (-2/7, +infinite).
In the case of concavity down,
[tex]\begin{gathered} f^{\prime}^{\prime}(x)<0 \\ \Rightarrow2e^{7x}+7x*e^{7x}<0 \\ \Rightarrow2+7x<0 \\ \Rightarrow-\frac{2}{7}>x \end{gathered}[/tex]
Thus, f(x) is concave down when x in (-infinite, -2/7).
The answer is the fifth and last option (top to bottom).
