# Linear Approximation

• May 2nd 2009, 03:37 PM
sk8erboyla2004
Linear Approximation
The problem given was $f(x)=e^{x^2}$ near x = 5 solve for exact answer

$e^{x^2}+2e^{x^2}(x-5)$

$3e^{x^2}(x-5)$ and then plug in 5 for x

$3e^{5^2}(x-5)$

simplify

$3xe^{25}-15ex^25$

is this fine?
• May 2nd 2009, 05:42 PM
Scott H
If the expression were $e^{x^2}(x-5)+2e^{x^2}(x-5)$, then you could derive $3e^{x^2}(x-5)$, but not in this case.
• May 2nd 2009, 06:22 PM
sk8erboyla2004
• May 2nd 2009, 06:38 PM
redsoxfan325
Quote:

Originally Posted by sk8erboyla2004
The problem given was $f(x)=e^{x^2}$ near x = 5 solve for exact answer

$e^{x^2}+2e^{x^2}(x-5)$

$3e^{x^2}(x-5)$ and then plug in 5 for x

$3e^{5^2}(x-5)$

simplify

$3xe^{25}-15ex^25$

is this fine?

First of all, the derivative of $e^{x^2}$ is $2xe^{x^2}$, not $2e^{x^2}$

The formula you want is $f(5)+f'(5)(x-5) = e^{25}+ 10e^{25}(x-5) = 10e^{25}x-50e^{25}+e^{25} = \boxed{10e^{25}x-49e^{25}}$

This won't be a great approximation though, since $e^{x^2}$ increases very rapidly.
• May 2nd 2009, 06:52 PM
sk8erboyla2004
Thanks mate!

would leaving

$e^25+10e^25(x-5)$

be sufficient?
• May 2nd 2009, 06:55 PM
redsoxfan325
Quote:

Originally Posted by sk8erboyla2004
Thanks mate!

would leaving

$e^{25}+10e^{25}(x-5)$

be sufficient?

I don't know how your grader will prefer it to be written, but it is still the correct answer; it's just up to personal preference how you write it. So I would say it's still fine.
• May 2nd 2009, 07:31 PM
sk8erboyla2004
Thanks once again!