Canculus 1 problem?

Question

What is the equation of a tangent line if y=(x^2-1)/(x^2+1) and (0,-1)?

I found y' first which i got 4x/(x^2+1)^2
But, when i put 0 into the function, i got the slope of 0.
So, the final equation became y+1=0.
And, it does not seem to me the answer...


Can you plz check and tell me how to reach the right answer?

Thanks :)

Answer 1

ur answer is correct. the derivative of the curve at the point (0,-1) is zero which means that the tangent at that point would be a straight horizontal line. so the equation of the line is y=-1

Answer 2

Actually, you have the right answer. Note that we can rewrite y as
y = 1 - [2 / (x^2 + 1)]

Since x^2 + 1 is always positive with a minimum (of 1) at x = 0, 2 / (x^2 + 1) is always positive with a maximum (of 2) at x = 0, and therefore y(x) has a minimum of -1 at x = 0. Since y has a minimum we expect the derivative to be 0 and the tangent line to be horizontal, just as you have found.

Answer 3

y=(x^2-1)/(x^2+1)
y'=[(x^2+1)(2x)-(x^2-1)(2x)]/(x^2+1)^2
find the slope at P(0,-1)
y'=m
y'=[(0^2+1)(2*0)-(0^2-1)(2*0)]/(0^2+1)^2
y'=0
using point slope form to determine
the equation of the line
y-y1=m(x-x1)
y-1=0(x-(-1))
y-1=0