Question

find the value of x at which the function has a possible relative maximum or minimum point. (recall that $e^{x}$ is for all x.) use the second derivative to determine the nature of the function at this point.
$f(x)=(5 x - 3) e^{3 - 6 x}$
what are the first and second derivatives of $f(x)$?
$f^{prime}(x)=e^{3 - 6 x}(-30 x + 23)$
$f^{prime prime}(x)=e^{3 - 6 x}(180 x - 168)$
the function has a relative point at $x=$ (simplify your answer. type)

Explanation:

Step1: Find critical points

Set \(f^{\prime}(x)=0\). Since \(e^{3 - 6x}>0\) for all \(x\), we solve \(-30x + 23=0\).
\(-30x+23 = 0\)
\(-30x=-23\)
\(x=\frac{23}{30}\)

Step2: Use the second - derivative test

Substitute \(x = \frac{23}{30}\) into \(f^{\prime\prime}(x)\).
\(f^{\prime\prime}(x)=e^{3 - 6x}(180x - 168)\)
\(f^{\prime\prime}(\frac{23}{30})=e^{3-6\times\frac{23}{30}}(180\times\frac{23}{30}-168)\)
First, simplify the exponent: \(3-6\times\frac{23}{30}=3-\frac{23}{5}=\frac{15 - 23}{5}=-\frac{8}{5}\)
Second, simplify the polynomial part: \(180\times\frac{23}{30}-168 = 138 - 168=-30\)
So \(f^{\prime\prime}(\frac{23}{30})=e^{-\frac{8}{5}}\times(- 30)<0\)

Answer:

The function has a relative maximum point at \(x = \frac{23}{30}\)