Question

let $f$ be the function given by $f(x)=2cos x + 3e^{x}$. what is the value of $f(0)$?
a - 3
b 0
c 3
d 5

Explanation:

Step1: Differentiate term - by - term

The derivative of $\cos x$ is $-\sin x$ and the derivative of $e^{x}$ is $e^{x}$. Using the sum rule of differentiation $(u + v)'=u'+v'$, if $f(x)=2\cos x + 3e^{x}$, then $f'(x)=2(-\sin x)+3e^{x}=-2\sin x + 3e^{x}$.

Step2: Evaluate at $x = 0$

Substitute $x = 0$ into $f'(x)$. We know that $\sin(0)=0$ and $e^{0}=1$. So $f'(0)=-2\sin(0)+3e^{0}=-2\times0 + 3\times1=3$.

Answer:

C. 3