# Some Derivatives Questions

1. Jun 20, 2011

### StopWatch

1. The problem statement, all variables and given/known data

Let f(x) = axe^((bx)^2). Find the value for a times b if it's known that there's a max value of 2 at x = 3.

Second, There is one line which is tangent to the curve y = 1/x, at some point A and at the same time tangent to the curve x^2 at some point B. What is the distance between
A and B?

2. Relevant equations

For the first I realize that I have f'(3) = 0 and (I think) f(3) = 2.

Second, I think I can set the derivatives equal to each other and solve for x.

3. The attempt at a solution

I can then use these equations to solve for a and b, however I guess this is more an algebra problem because I seem unable to do so. I also need verification that that second equation makes sense. If it does, I think I'll be more able to do this question because I can solve for b in a way I couldn't with the first equation (if I took ln of both sides ln0 would be undefined and it wouldn't make sense, I think).

Second, when I tried to solve for x I ended up with x = 1/2 which lead me to a solution eventually, by the distance formula, of 9/16 + 1/16 which is not correct based on the test answers.

Any help would be appreciated here, thanks guys!

2. Jun 20, 2011

### lanedance

$$f(x) = axe^{(bx)^2}$$
$$f(3) = 2 = a3e^{(b3)^2} = 27ae^{b^2}$$

3. Jun 20, 2011

### lanedance

for the 2nd the tangents will be parallel, but it states a single line will also pass through both points, so you've got a little more info than just the tangent direction. you can describe the line by a gradient (tangent direction) and an intercept

4. Jun 20, 2011

### lanedance

note or 1 that you'll need to apply the chain rule (and product rule)
$$\frac{d}{dx}g(f(x)) = g'(f(x))f'(x)$$

then i think the exponential part should cancel

5. Jun 20, 2011

### StopWatch

I'm sorry about taking so long to respond to my own thread, I really appreciate the help but some things came up around here that I had to take care of.

f(3) = 2 = a3e^{(b3)^2} = 27ae^{b^2} - I don't even quite know why this is true.

The problem I have with the derivative is that I'll get f'(3) = 0 = a(e^b3^2) + a3(e^b3^2)(6b) = (a + a3)(e^b3^2) = a4(e^b3^2), which I don't know how to solve further because anything I put on the other side will end up being zero. Should I have instead put that a(e^b3^2) = -(a3(e^b3^2)(6b)) or?

6. Jun 20, 2011

### lanedance

7. Jun 20, 2011

### lanedance

not too sure what you're doing here , but you have 2 equations and 2 constants so can solve, assuming there is a solution
f(3) = 2 = 3ae^{(b3)^2}
f'(3) = 0 = a(e^b3^2) + a3(e^b3^2)(6b) = (e^b3^2)(a+3a6b)

clearly form the first, a is non-zero
2/(3a) = e^{(b3)^2}

substitute this into the 2nd