[SOLVED] curve sketching problem

xxlvh · #1 $Old$ November 5th, 2009, 09:40 PM

Consider the function $f(x) = e^{arctan(2x)}$

1. Find the horizontal asymptotes of the function.
2. State the interval(s) where the function is increasing.
3. Find the inflection point.
4. Find the intervals where the graph is concave up and concave down.

It appears I've ran into a problem here. All I could do on my own so far was find f'(x):
$f'(x) = e^{arctan(2x)}(\frac{2}{1+4x^2}) = \frac{2e^{arctan(2x)}}{1+4x^2}$

Setting this equal to 0, $e^{arctan(2x)} = 0$ which has no solution? Doesn't look quite right to me.

Defunkt · #2 $Old$ November 6th, 2009, 12:48 AM

That's correct -- the function is monotonic over $\mathbb{R}$

xxlvh · #3 $Old$ November 6th, 2009, 09:39 PM

Ah, thank you! I should have noticed that just by looking at the function itself.

I'm still not completely positive on whether or not I have done the rest of the problem correctly.
This is what I have for the second derivative:

$f''(x) = \frac{(1+4x^2)\frac{4e^{arctan(2x)}}{(1+4x^2)} - (2e^{arctan(2x)})(8x)}{(1+4x^2)^2}$

Which then simplifies to be:

$f''(x) = (\frac{2e^{arctan(2x)}}{(1+4x^2)^2})(2-8x)$

I suppose the only way for it to equal zero would be x=1/4
So the coordinates of the point of inflection are $(\frac{1}{4}, e^{arctan(1/2)})$

And it would be concave up on $(-\infty,\frac{1}{4})$ and concave down on $(\frac{1}{4},\infty)$

EDIT: Already submitted this answer and it turned out to be correct