$f(x) = e^x$ and $a = 1$. Find the linear approximation $L(x)$

Question

I a little confused on this question and I feel I shouldn't be. So, I take the derivative of f(x) which is $f'(x)=e^x$

Next I plug in the point $a = 1$, which then gives me the slope $2.71$

Knowing $L(x)= f'(a)(x-a)+ f(a)$

I plug $f(1)$ into the original function which is just $e^x = 2.71$

So, now using the $L(x)$ equation of the line. I get the following:

$L(x) = 2.71(x-1) + 2.71$

Which can further simplify to $2.71x$

I'm I going down the correct path with this problem or I'm overlooking something?

Answer 1

Yes you are correct.

$f(x_0+h)=f(x_0)+f'(x_0)\cdot h+o(h)$

$f'(x)=e^x$

Now set $x_0$=1:

$f(1+h)=e^1+e^1\cdot h +o(h) = e + eh +o(h)$