Hello (again), please help me with this problem. I don't really know where to being. Thanks.

http://i13.photobucket.com/albums/a2...f?t=1262653181

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- Jan 4th 2010, 04:06 PMpocketasianA Tangent to a Graph
Hello (again), please help me with this problem. I don't really know where to being. Thanks.

http://i13.photobucket.com/albums/a2...f?t=1262653181 - Jan 4th 2010, 04:24 PMpickslides
start by making $\displaystyle f'(x) = 3$

- Jan 4th 2010, 04:27 PMdedust
- Jan 4th 2010, 04:39 PMpocketasian
Alright, thanks. Since there's an x as an exponent, how should I go about solving for it? Should I take the natural log of each side? And if I do, where do I go from there?

- Jan 4th 2010, 05:37 PMmr fantastic
Use the chain rule to differentiate the given function. If you cannot do this, I suggest you review the technique and appropriate examples from your classnotes and textbook. Note that the correct answer for the derivative has already been given in an earlier post.

- Jan 4th 2010, 05:44 PMpocketasian
Yes, I know how to solve for the derivative. I just need a push with how to solve for x after I differentiate the function. My algebra is a bit rusty, but I do remember something about taking the natural log of each side to bring the x in the exponent to the front...or something like that.

- Jan 4th 2010, 05:47 PMSudharaka
Dear friends,

I had thought of this problem for some time. I think this curve does not contain a point such that the tangent is 3. Please see the attachment I have given below. I any of you find the attachment unreadable or any problem with my solution please reply me.

Hope this helps. - Jan 4th 2010, 05:48 PMmr fantastic
- Jan 4th 2010, 05:54 PMskeeter
why did you take the derivative twice?

$\displaystyle f'(x) = 3$

$\displaystyle 16xe^{4x^2} = 3$

$\displaystyle x \approx 0.1676$ - Jan 4th 2010, 05:56 PMmr fantastic
- Jan 4th 2010, 05:59 PMpocketasian
0.168 is the correct answer, which is what skeeter came up with. Thanks everyone for helping me solve this problem. It actually makes better sense now.