Find all points where f(x) has a horiztonal tangent line

SPhy
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1. Given.

F(x) = 4x/(x^2+1)

2. Problem

Find all points (x,y) where the function has a horiztonal tangent line

3. Attempt

I took the derivative of F(x) and came to

(-4x^2+4)/(x^2+1)^2

I set it equal to zero and found an x value of 1. I used that x value and plugged it into the original function, so f(1), and got a Y value of 2, but according to my professor this was not correct. However I asked him this after the test. It was a test question and on the test I stated at no point does the function have a horizontal tangent line.

I figured I was wrong because after the test I redid the problem and came to (1,2). Although as stated, my professor said (1,2) was incorrect. I haven't got my test back, so maybe there was no solution and I did the problem correctly originally, but I think there is a solution.

Thanks!
 
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SPhy said:
1. Given.

F(x) = 4x/(x^2+1)

2. Problem

Find all points (x,y) where the function has a horiztonal tangent line

3. Attempt

I took the derivative of F(x) and came to

(-4x^2+4)/(x^2+1)^2

I set it equal to zero and found an x value of 1. I used that x value and plugged it into the original function, so f(1), and got a Y value of 2, but according to my professor this was not correct. However I asked him this after the test. It was a test question and on the test I stated at no point does the function have a horizontal tangent line.

I figured I was wrong because after the test I redid the problem and came to (1,2). Although as stated, my professor said (1,2) was incorrect. I haven't got my test back, so maybe there was no solution and I did the problem correctly originally, but I think there is a solution.

Thanks!

There are two points where the function has a horizontal tangent. You've got one. Can you find the other one?
 
Dick said:
There are two points where the function has a horizontal tangent. You've got one. Can you find the other one?

Hmm I think I forgot to consider the +/- on the square root. So the other point would be (1,-2)?
 
SPhy said:
Hmm I think I forgot to consider the +/- on the square root. So the other point would be (1,-2)?

Yes, that's what you forgot. And I'm assuming you meant (-1,-2) and that's just a typo.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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