Why does this work?

[RONJOHNSON]

For homework, I was assigned a problem that read,

Find what the value of "a" is given that the tangent equation of f(x) = $$\frac{a}{(x+2)^2}$$ is y = -2x+ 8.

What I did was:

$$f'(x)=\frac{-2a}{(x+2)^3}$$ and then $$\frac{a}{(x+2)^2}$$=$$\frac{-2a}{(x+2)^3}$$
After simplifying that, I got x = -4.

I plugged that into y = -2x + 8 and got 16.

Then, since the point of tangency is equal to the tangent line and the original graph, $$16 =\frac{a}{(0+2)^2}$$.

This, of course, meant a = 64.

However, when we went of this in class, I found out that I was correct, but the teacher showed us a different way to do it. He set the deviate equal to the slope (-2), and then the tangent line with the original function, since x should have the same value when the y is the same. When you do a system of equations, you get x = 2. Plugging that into y = -2x + 8 gives you 4. Since this occurs at the same place as the original function, you can say $$4 = \frac{a}{(2+2)^2}$$. This, of course, also gives you a = 64.

My question is how come my first method worked. I don't see how I could set the original function equal with the derivative, and then plug that into the tangent line in a place that's not even the point of tangency?

Thank you!

 

[lev888]

Tangent at what point?

 

[RONJOHNSON]

There was no point given, the idea is that we have to pick the value of "a" that would make the original graph tangent to the tangent line.

 

[Dr.Peterson]

My question is how come my first method worked. I don't see how I could set the original function equal with the derivative, and then plug that into the tangent line in a place that's not even the point of tangency?

Even a clock that has stopped is correct twice a day! This may be as random as that.

Since your "method" makes no sense, I don't think there's much value in trying to see why it happened to give the correct answer in this case.

Did you have some reason to do what you did?

 

[BigBeachBanana]

You got lucky where the derivative and its function intercept at a point that is also the tangent line of some unknown point. If the intersecting point didn't lie on the tangent line, your method would not work. I would buy a lottery ticket with this much luck.

 

[RONJOHNSON]

Even a clock that has stopped is correct twice a day! This may be as random as that.

Since your "method" makes no sense, I don't think there's much value in trying to see why it happened to give the correct answer in this case.

Did you have some reason to do what you did?

I did actually, but looking back on it, it was flawed. My thinking was that when f(x) and f'(x) are equal to each other, the slope and the y-coordinate would be the same, as the original graph and the derivative would be the same at that x. And while this logic is not flawed, plugging that "x" into the tangent line equation doesn't make much sense (since it is not at the point of tangency), but I just did it because I didn't know what else to do. I get that the f(x) and the f'(x) intersect at x = -4, but I don't understand how the tangent line intersects their as well (which it does, otherwise it wouldn't work). The weird thing is that when I changed the numbers a little, the same thing ended up happening.

 

[RONJOHNSON]

You got luck where the derivative and its function intercept at a point on the tangent line of some unknown point. I would buy a lottery ticket with this much luck.

Haha, the weird thing is that I changed the numbers very slightly and the same thing happened. I wonder if it has something to do with the fact that we are defining the original function through the derivative, or if it just happened again, or if there is some weird property in those equations. Either way, I got really lucky. I'm not sure why I said homework, because it was actually a quiz. So I guess I really lucked out!

 

[BigBeachBanana]

Haha, the weird thing is that I changed the numbers very slightly and the same thing happened. I wonder if it has something to do with the fact that we are defining the original function through the derivative, or if it just happened again, or if there is some weird property in those equations. Either way, I got really lucky. I'm not sure why I said homework, because it was actually a quiz. So I guess I really lucked out!

Have you looked at the function and the derivative graphs? They're almost on top of each other, so if you picked a number slightly close to it, it would probably still intersect.