Question

question
find the equation of the line tangent to the graph of
$f(x)=3\sin(x)-2\cos(x)$ at $x = -\pi$
(enter an exact answer.)
provide your answer below:
$y = $

Explanation:

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

Using the derivative rules $\frac{d}{dx}(\sin x)=\cos x$ and $\frac{d}{dx}(\cos x)=-\sin x$, we have $f'(x)=3\cos(x)+ 2\sin(x)$.

Step2: Evaluate $f(-\pi)$

$f(-\pi)=3\sin(-\pi)-2\cos(-\pi)=3\times0 - 2\times(-1)=2$.

Step3: Evaluate $f'(-\pi)$

$f'(-\pi)=3\cos(-\pi)+2\sin(-\pi)=3\times(-1)+2\times0=-3$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-\pi,2)$ and $m = f'(-\pi)=-3$. So $y - 2=-3(x+\pi)$.

Step5: Rearrange to slope - intercept form

$y=-3x-3\pi + 2$.

Answer:

$y=-3x-3\pi + 2$