this is similar to section 3.4 problem 24: for the function (k(x) = 2xe^{3x}), do the following: (a)…

Explanation:

Step1: Find the derivative of $K(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 2x$ and $v = e^{3x}$. $u^\prime=2$ and $v^\prime = 3e^{3x}$. Then $K^\prime(x)=2e^{3x}+2x\times3e^{3x}=2e^{3x}(1 + 3x)$.

Step2: Find the critical numbers

Set $K^\prime(x)=0$. Since $e^{3x}>0$ for all real $x$, we solve $1 + 3x=0$. So $x=-\frac{1}{3}$.

Step3: Apply the First - Derivative Test

Choose a test point to the left of $x =-\frac{1}{3}$, say $x=- \frac{2}{3}$. Then $K^\prime(-\frac{2}{3})=2e^{- 2}(1-2)=-2e^{-2}<0$. Choose a test point to the right of $x =-\frac{1}{3}$, say $x = 0$. Then $K^\prime(0)=2e^{0}(1 + 0)=2>0$. Since the first - derivative changes sign from negative to positive at $x =-\frac{1}{3}$, the function has a relative minimum at $x=-\frac{1}{3}$.

Answer:

(a) $x_{c}=-\frac{1}{3}$ (b) Near the critical number $x_{c}=-\frac{1}{3}$, when $x<x_{c}$, the first derivative is negative, and when $x>x_{c}$, the first derivative is positive. Therefore the function has a relative minimum at $x_{c}=-\frac{1}{3}$.