let f be the function given by f(x)=2cos x + 3e^x. what is the value of f(0)?

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 $f'(x)=- 2\sin x+3e^{x}$.

Step2: Evaluate at $x = 0$

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

Step3: Calculate the result

Since $\sin(0) = 0$ and $e^{0}=1$, we have $f'(0)=-2\times0 + 3\times1=3$.

Answer:

$3$