Find the points on the curve of y=x/(x^2+1) where the tangent line is horizontal.

Answer 1

To get horizontal points on a graph means to find zeroes of the derivative

We need the quotient rule. "bottom top prime minus top bottom prime, over bottom squared"

(x²+1)(1)-(x)(2x) / (x²+1)²

(x²+1 - 2x²)/(x²+1)²

(1-x²)/(x²+1)²

Now, the bottom is always positive, so we need not worry about /0 errors. We just need to check the top for zeroes.

1-x²=0

Clearly, x=±1 are the solutions

Putting x=1, we get (1)/(1+1) = 1/2

Putting x=-1, we get(-1)/(1+1) = -1/2

Solutions are (-1, -1/2),(1/1/2) ■