# Linear Approximation

• May 2nd 2009, 02:37 PM
sk8erboyla2004
Linear Approximation
The problem given was \$\displaystyle f(x)=e^{x^2}\$ near x = 5 solve for exact answer

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

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

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

simplify

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

is this fine?
• May 2nd 2009, 04:42 PM
Scott H
If the expression were \$\displaystyle e^{x^2}(x-5)+2e^{x^2}(x-5)\$, then you could derive \$\displaystyle 3e^{x^2}(x-5)\$, but not in this case.
• May 2nd 2009, 05:22 PM
sk8erboyla2004
• May 2nd 2009, 05:38 PM
redsoxfan325
Quote:

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

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

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

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

simplify

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

is this fine?

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

The formula you want is \$\displaystyle 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 \$\displaystyle e^{x^2}\$ increases very rapidly.
• May 2nd 2009, 05:52 PM
sk8erboyla2004
Thanks mate!

would leaving

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

be sufficient?
• May 2nd 2009, 05:55 PM
redsoxfan325
Quote:

Originally Posted by sk8erboyla2004
Thanks mate!

would leaving

\$\displaystyle 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, 06:31 PM
sk8erboyla2004
Thanks once again!