Answer
\[\begin{align}
& a.\text{ }x=-\frac{1}{2} \\
& b.\text{ the tangent line to the graph }f\left( x \right)=x{{e}^{2x}}\text{ } \\
& \text{at }x=-\frac{1}{2}\text{ is horizontal} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \text{Let }f\left( x \right)=x{{e}^{2x}} \\
& \text{a}\text{.} \\
& \text{Calculate }f'\left( x \right) \\
& f'\left( x \right)=\frac{d}{dx}\left[ x{{e}^{2x}} \right] \\
& f'\left( x \right)=x\frac{d}{dx}\left[ {{e}^{2x}} \right]+{{e}^{2x}}\frac{d}{dx}\left[ x \right] \\
& f'\left( x \right)=x\left( 2{{e}^{2x}} \right)+{{e}^{2x}}\left( 1 \right) \\
& f'\left( x \right)=2x{{e}^{2x}}+{{e}^{2x}} \\
& \text{Let }f'\left( x \right)=0\text{ to find the values of }x\text{ at which the slope of} \\
& \text{the curve }f\left( x \right)\text{ is 0}\text{.} \\
& 2x{{e}^{2x}}+{{e}^{2x}}=0 \\
& {{e}^{2x}}\left( 2x+1 \right)=0 \\
& {{e}^{2x}}=0\text{ or }2x+1=0 \\
& {{e}^{2x}}\text{ is always }>\text{0 for all real number }x,\text{ then} \\
& 2x+1=0 \\
& x=-\frac{1}{2} \\
& \text{The value at which the derivative of the function is }0\text{ is }x=-\frac{1}{2}. \\
& \\
& \text{b}\text{. } \\
& \text{From the graph of }f\left( x \right)\text{ shown below we can say that the} \\
& \text{line tangent to the graph }f\left( x \right)=x{{e}^{2x}}\text{ at }x=-\frac{1}{2}\text{ is horizontal.} \\
& \text{Graph } \\
\end{align}\]
