$$f(x) = erf$$ is a function which satisfies f(1) = erf (equal approx) 0.84270 and
$$f'(x) = 2 [(e ^{-x ^{2}})/(\sqrt{\pi})]$$

use a linear approximation to approximate f(0.9):

Question

$$f(x) = erf$$ is a function which satisfies f(1) = erf (equal approx) 0.84270 and
$$f'(x) = 2  [(e ^{-x ^{2}})/(\sqrt{\pi})]$$

use a linear approximation to approximate f(0.9):

anonymous · Answer

I know the equation for a linear approximenation is l(x) = f(a) + f'(a)(x-a) but I'm confused on how to insert the stuff

mr.math · Answer

It's nothing but direct substitution with x=0.9, a=1.

anonymous · Answer

Ah, I guess the only thing I'm really confused on is the f, and for f(a) do I just put 0.9 in f'(x) and whatever is the result is then f(a)?

mr.math · Answer

$a=1 \implies f(a)=f(1)= 0.84270$.

mr.math · Answer

$$f'(a)=f'(1)=2\frac{e^{-1}}{\sqrt{\pi}}.$$

mr.math · Answer

Makes sense?

anonymous · Answer

Yeah, I get it now, thanks for your help. I appreciate it :D

mr.math · Answer

You're welcome :)