# Thread: A Tangent to a Graph

1. ## A Tangent to a Graph

2. start by making $\displaystyle f'(x) = 3$

3. Originally Posted by pocketasian
slope = $\displaystyle \frac{d(f(x))}{dx} = 3$

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

solve for x

4. 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?

5. 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.

6. 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.

7. 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.

8. 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.

9. why did you take the derivative twice?

$\displaystyle f'(x) = 3$

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

$\displaystyle x \approx 0.1676$

10. 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.

11. 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.