Question

let f be the function given by f(x)=3 sin x + 8e^x. what is the value of f(0)?

Explanation:

Step1: Differentiate term - by - term

The derivative of \(y = 3\sin x\) is \(y'=3\cos x\) using the rule \(\frac{d}{dx}(\sin x)=\cos x\), and the derivative of \(y = 8e^{x}\) is \(y' = 8e^{x}\) using the rule \(\frac{d}{dx}(e^{x})=e^{x}\). So, \(f'(x)=3\cos x + 8e^{x}\).

Step2: Evaluate at \(x = 0\)

Substitute \(x = 0\) into \(f'(x)\). We know that \(\cos(0)=1\) and \(e^{0}=1\). Then \(f'(0)=3\cos(0)+8e^{0}=3\times1 + 8\times1\).

Step3: Calculate the result

\(f'(0)=3 + 8=11\).

Answer:

D. 11