this is section 3.3 problem 24: for the function f(x)=xe^{3x}, determine its inflection point(s) if exists…

Explanation:

Step1: Find the first - derivative

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

Step2: Find the second - derivative

Again use the product rule on $f^\prime(x)=e^{3x}(1 + 3x)$. Let $u = 1+3x$ and $v = e^{3x}$, $u^\prime=3$, $v^\prime = 3e^{3x}$. $f^{\prime\prime}(x)=3e^{3x}+3e^{3x}(1 + 3x)=3e^{3x}(1+(1 + 3x))=3e^{3x}(2 + 3x)$.

Step3: Set the second - derivative equal to zero

Since $e^{3x}\gt0$ for all real $x$, we set $2 + 3x=0$. $3x=-2$, so $x=-\frac{2}{3}\approx - 0.67$.

Step4: Find the $y$ - value

Substitute $x =-\frac{2}{3}$ into the original function $f(x)=xe^{3x}$. $f(-\frac{2}{3})=-\frac{2}{3}e^{3\times(-\frac{2}{3})}=-\frac{2}{3}e^{-2}=-\frac{2}{3e^{2}}\approx-\frac{2}{3\times7.39}\approx - 0.09$.

Answer:

$(x_1,y_1)=(-0.67,-0.09)$