# A Tangent to a Graph

• Jan 4th 2010, 04:06 PM
pocketasian
A Tangent to a Graph
http://i13.photobucket.com/albums/a2...f?t=1262653181
• Jan 4th 2010, 04:24 PM
pickslides
start by making $\displaystyle f'(x) = 3$
• Jan 4th 2010, 04:27 PM
dedust
Quote:

Originally Posted by pocketasian
http://i13.photobucket.com/albums/a2...f?t=1262653181

slope = $\displaystyle \frac{d(f(x))}{dx} = 3$

$\displaystyle \frac{d(f(x))}{dx} = 16xe^{4x^2} = 3$

solve for x
• Jan 4th 2010, 04:39 PM
pocketasian
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 PM
mr fantastic
Quote:

Originally Posted by pocketasian
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?

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 PM
pocketasian
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 PM
Sudharaka
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 PM
mr fantastic
Quote:

Originally Posted by pocketasian
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.

$\displaystyle 16 x e^{4x^2} = 3$ cannot be solved in closed form using algebra. You're undoubtedly expected to use technology to get an approximate solution to a given number of decimal places.
• Jan 4th 2010, 05:54 PM
skeeter
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 PM
mr fantastic
Quote:

Originally Posted by Sudharaka
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.

Since f'(0) = 0 and the function increases without bound as x --> +oo it's not difficult to see that a solution to f'(x) = 3 does exist. In fact, the solution lies between x = 0 and x = 1.
• Jan 4th 2010, 05:59 PM
pocketasian
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.