ScholMaths
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- Joined
- Apr 30, 2012
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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 nth derivative
Hence, nature of stationary values at f(x)=+-1 are indeterminate.
But how do I show -1<=f(x)<=1?
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 nth derivative
Hence, nature of stationary values at f(x)=+-1 are indeterminate.
But how do I show -1<=f(x)<=1?