Question

find an equation of the line tangent to the graph of g(x)=3e^(-7x) at the point (0,3). the equation of the line is y =

Explanation:

Step1: Find the derivative of G(x)

Using the chain - rule, if $y = 3e^{-7x}$, let $u=-7x$, then $y = 3e^{u}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=3e^{u}$, and the derivative of $u$ with respect to $x$ is $\frac{du}{dx}=-7$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $G^\prime(x)=3e^{-7x}\cdot(-7)=-21e^{-7x}$.

Step2: Evaluate the derivative at x = 0

Substitute $x = 0$ into $G^\prime(x)$. $G^\prime(0)=-21e^{-7\times0}=-21\times1=-21$. The value of the derivative at $x = 0$ is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(0,3)$ and $m=-21$. Substituting these values, we get $y - 3=-21(x - 0)$.

Step4: Simplify the equation

$y-3=-21x$, so $y=-21x + 3$.

Answer:

$y=-21x + 3$