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Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 4.2 : Critical Points
14. Determine the critical points of \(f\left( x \right) = 5x\,{{\bf{e}}^{9 - 2x}}\).
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Start SolutionWe’ll need the first derivative to get the answer to this problem so let’s get that.
\[f'\left( x \right) = 5{{\bf{e}}^{9 - 2x}} + 5x\left( { - 2} \right){{\bf{e}}^{9 - 2x}} = 5{{\bf{e}}^{9 - 2x}}\left( {1 - 2x} \right)\]We did some quick factoring to help with the next step and while it doesn’t technically need to be done it will significantly reduce the amount work required in the next step.
Show Step 2Recall that critical points are simply where the derivative is zero and/or doesn’t exist.
This derivative exists everywhere and so we don’t need to worry about that. Therefore, all we need to do is determine where the derivative is zero.
Notice as well that because we know that exponential functions are never zero and so the derivative will only be zero if,
\[1 - 2x = 0\hspace{0.25in} \to \hspace{0.25in}\,x = {\textstyle{1 \over 2}}\]So, we have a single critical point, \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = {\textstyle{1 \over 2}}}}\), for this function.