Help with sketching f'=1-f^2 and showing -1<=f<=1

[ScholMaths]

A function satisfies the D.E. f'(x)=1-[f(x)]² and f(0)=0. How can I sketch f and show -1<=f(x)<=1 without solving the D.E.?
This is from an old exam paper I'm working through. I have the following working:

1) f'(0)=1-0=1. Hence, f passes through the origin with gradient 1

2) f'' = -2ff' = 2f(f²-1) => f''(0)=0. Hence there is an increasing point of inflexion at (0,0)

3) For stationary values, f'=0 => f(x)=1 or -1. But f''=0 for f(x)=1 or -1. Furthermore, fⁿ=0 for f(x)=1 or -1 where n is the n^th derivative
Hence, nature of stationary values at f(x)=+-1 are indeterminate.

But how do I show -1<=f(x)<=1?

 

[HallsofIvy]

If \(\displaystyle f'(x)= 1- f^2(x)= (1- f(x))(1+ f(x))\) then f'= 0 wherever f is 1 or -1. The constant functions f(x)= 1 and f(x)= -1 are solutions to the differential equation. That, in turn, means that, by the "existence and uniqueness theorem" for such differential equations, the graph of y= f(x) cannot cross the lines y= 1 and y= -1. y= 1 and y= -1 are "equilibrium" solutions. You can, in fact, see that if f(x)< -1, then 1- f(x) is positive while 1+ f(x) is negative so that y' is negative. That, if, for some x, f(x)< -1 then f decreases away from -1 as x increases. If -1< f(x)< 1 then both 1- f(x) and 1+ f(x) are positive so y' is positive. If, for some x, f(x) is close to -1 but larger, then f(x) increases away from -1. f(x)= -1 is an unstable equilibrium solution. Similarly, you can show that f(x)= 1 is a stable equilibrium solution.

Yes, the graph of y= f(x) with f satisfying \(\displaystyle f'(x)= 1- f^2(x)\) and f(0)= 0 crosses through (0,0) at a slope of 45 degrees. As x increases, it then approaches y= 1 as an asymptote, getting arbitrarily close to y=1 but never reaching it. As x decreases from x= 0, y= f(x) approaches -1 as an asymptote.