$f(x) = e^x$ 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?

You should use $e$ instead of $2.71$ but you are on the correct path — stity, Nov 17 '15 at 00:21

score 0 · Answer 1 · answered Nov 17 '15 at 00:25

0

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:

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

answered Nov 17 '15 at 00:25

Guy

Ok, this is good but here is where I get confused the next part to the question states "Use L(x) to estimate f(0.7)". So, I can plug 0.7 into 2.71x, right? and I get 1.897. Which isn't close to e^0.7 = 2.01 – user3339882 Nov 17 '15 at 00:38
2

True.
The closer you evaluate your approximation to $x_0=1$, the better it will be. The approximation using the derivative is a local property. With differentiable functions, the more you zoom into a point, it behaves more linearly in its neighbourhood.

link

This visualization should clarify it.
– Guy Nov 17 '15 at 00:47

1 Answers1