Question

The function g is given by g(x)=eax+f(x) for all real numbers, where a is a constant.
Find g ′(0) and g ″(0) in terms of a. Show the work that leads to your answers.

Answer 1

I think that you left out some conditions from the questions.

Let's suppose that,
\(\color{#000000 }{ \displaystyle f'(0) =-5 }\)
\(\color{#000000 }{ \displaystyle f''(0) =1 }\)

Then, you can differentiate the function and get:
\(\color{#000000 }{ \displaystyle g'(x)=ae^{ax}+f'(x) }\)
\(\color{#000000 }{ \displaystyle g''(x)=a^2e^{ax}+f''(x) }\)

Now, let's plug in \(\small x=0\) to solve for \(\small f'(0)\) and \(\small f''(0)\).
\(\color{#000000 }{ \displaystyle g'(0)=a+f'(0) }\)
\(\color{#000000 }{ \displaystyle g''(0)=a^2+f''(0) }\)

Now, you can plug in the given values for \(\small f'(0)\) and \(\small f''(0)\).
\(\color{#000000 }{ \displaystyle g'(0)=a-5 }\)
\(\color{#000000 }{ \displaystyle g''(0)=a^2+1 }\)

Answer 2

I thought of using greek letters, but I threw out the fenciness to obtain understanding