Question

an equation of the normal line to the graph of f(x)=3xe^x + 5 at x = 0 is
○ y=-3x + 5
○ y = 3x+5
○ y=-\frac{1}{3}x + 5
○ y=\frac{1}{3}x + 5

Explanation:

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

Use product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 3x$ and $v=e^{x}$. The derivative of $u = 3x$ is $u^\prime=3$, and the derivative of $v = e^{x}$ is $v^\prime=e^{x}$. So $f^\prime(x)=(3x)^\prime e^{x}+3x(e^{x})^\prime=3e^{x}+3xe^{x}$.

Step2: Evaluate the derivative at $x = 0$

Substitute $x = 0$ into $f^\prime(x)$. $f^\prime(0)=3e^{0}+3\times0\times e^{0}=3\times1 + 0=3$.

Step3: Find the slope of the normal line

The slope of the normal line $m_{n}$ and the slope of the tangent line $m_{t}$ are related by $m_{n}=-\frac{1}{m_{t}}$. Since $m_{t}=f^\prime(0) = 3$, then $m_{n}=-\frac{1}{3}$.

Step4: Find the point on the curve at $x = 0$

Substitute $x = 0$ into $f(x)$: $f(0)=3\times0\times e^{0}+5 = 5$. The point is $(0,5)$.

Step5: Use the point - slope form to find the equation of the normal line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,5)$ and $m =-\frac{1}{3}$. So $y - 5=-\frac{1}{3}(x - 0)$, which simplifies to $y=-\frac{1}{3}x + 5$.

Answer:

$y=-\frac{1}{3}x + 5$