Question

find the value of x at which the function has a possible relative maximum or minimum point. (recall tha for all x.) use the second derivative to determine the nature of the function at this point.

$f(x)=(5x - 4)e^{1 - 2x}$

what are the first and second derivatives of $f(x)$?

$f(x)=e^{1 - 2x}(13 - 10x)$

$f(x)=e^{1 - 2x}(20x - 36)$

the function has a relative point at $x = $

(simplify your answer. type on.)

Explanation:

Step1: Find critical points

Set \(f^{\prime}(x) = 0\). Given \(f^{\prime}(x)=e^{1 - 2x}(13 - 10x)\), since \(e^{1 - 2x}>0\) for all \(x\), we solve \(13 - 10x = 0\).
\(13-10x = 0\Rightarrow10x=13\Rightarrow x=\frac{13}{10}=1.3\)

Step2: Use the second - derivative test

Substitute \(x = 1.3\) into \(f^{\prime\prime}(x)=e^{1 - 2x}(20x - 36)\)
First, when \(x = 1.3\), \(e^{1-2x}=e^{1 - 2\times1.3}=e^{-1.6}>0\)
Second, \(20x-36=20\times1.3 - 36=26 - 36=-10<0\)
So \(f^{\prime\prime}(1.3)=e^{1 - 2\times1.3}(20\times1.3 - 36)<0\)

Answer:

The function has a relative maximum point at \(x = 1.3\)