Answer
\[\begin{align}
& \text{Increasing on: }\left( -\infty ,\frac{1}{4} \right) \\
& \text{Decreasing on: }\left( \frac{1}{4},\infty \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=8x{{e}^{-4x}} \\
& \text{Calculating the first derivative} \\
& f'\left( x \right)=\frac{d}{dx}\left[ 8x{{e}^{-4x}} \right] \\
& f'\left( x \right)=8x\frac{d}{dx}\left[ {{e}^{-4x}} \right]+{{e}^{-4x}}\frac{d}{dx}\left[ 8x \right] \\
& f'\left( x \right)=8x\left( -4{{e}^{-4x}} \right)+{{e}^{-4x}}\left( 8 \right) \\
& f'\left( x \right)=-32x{{e}^{-4x}}+8{{e}^{-4x}} \\
& \text{Calculating the critical points, set }f'\left( x \right)=0 \\
& f'\left( x \right)=0 \\
& -32x{{e}^{-4x}}+8{{e}^{-4x}}=0 \\
& 8{{e}^{-4x}}\left( -4x+1 \right)=0 \\
& -4x+1=0 \\
& x=\frac{1}{4} \\
& \text{Set the intervals }\left( -\infty ,\frac{1}{4} \right),\left( \frac{1}{4},\infty \right) \\
& \text{Making a table of values} \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,\frac{1}{4} \right) & \left( \frac{1}{4},\infty \right) \\
\text{Test Value} & x=-1 & x=1 \\
\text{Sign of }f'\left( x \right) & \text{ }f'\left( -1 \right)>0 & \text{ }f'\left( 1 \right)<0 \\
\text{Conclusion} & \text{Increasing} & \text{Decreasing} \\
\end{matrix} \\
& \\
& \text{Summary} \\
& \text{Increasing on: }\left( -\infty ,\frac{1}{4} \right) \\
& \text{Decreasing on: }\left( \frac{1}{4},\infty \right) \\
\end{align}\]
