Question

Hi all,
I am currently working on 2B-1c and I am not sure where they are getting the constant c from. For this problem, they give us the derivative first and that y(0)=0 . There are no critical points and it looks like y'(x)>0 for all x. When I look at the endpoints of f'(x), I get 1/infinity for x approaching both negative and positive infinity. Without looking at the solution, I would never have thought to include this constant, c, as a horizontal asymptote. I guess my question is: where is this constant coming from?

Answer 1

y= f(x)
y' = f'(x) = 1/(1+x^2)

as x->infty, y' -> 0 , meaning the slope is approaching 0, and that means f(x) is approximately a horizontal line. i.e. as x-> infty, f(x) -> constant. This is the definition of an asymptote, y= c (c some constant)

we don't know what the constant is, but we know it's some non-zero value.

Answer 2

if we make a table of x and slope we have
x slope
-2 1/5
-1 1/2
0 1
1 1/2
2 1/5

at x=0 we are given y=0, so at point (0,0) we have a slope of +1
the slope is always positive, so we expect at x=1 to be at y > 0
and at x=2, to be at a larger y value than at x=1