Linear Approximation

**sk8erboyla2004** · May 2nd 2009, 02:37 PM

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?

**Scott H** · May 2nd 2009, 04:42 PM

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.

**sk8erboyla2004** · May 2nd 2009, 05:22 PM

So is my answer correct?

**redsoxfan325** · May 2nd 2009, 05:38 PM

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.

**sk8erboyla2004** · May 2nd 2009, 05:52 PM

Thanks mate!

would leaving

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

be sufficient?

**redsoxfan325** · May 2nd 2009, 05:55 PM

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.

**sk8erboyla2004** · May 2nd 2009, 06:31 PM

Thanks once again!

Thread: Linear Approximation

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