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