1. ## 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?

2. 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.

3. So is my answer correct?

4. 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.

5. Thanks mate!

would leaving

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

be sufficient?

6. 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.

7. Thanks once again!