Question

let f be the function given by f(x) = 2 cos 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