# Differentiability analysis

• Feb 22nd 2009, 10:35 AM
FractalMath
Differentiability analysis
How would I go about showing the existence of a differentiable funciton over the reals such that, for all x,

${\left( {f\left( x \right)} \right)^5} + f\left( x \right) + x = 0$?
• Feb 22nd 2009, 06:31 PM
HallsofIvy
Quote:

Originally Posted by FractalMath
How would I go about showing the existence of a differentiable funciton over the reals such that, for all x,

${\left( {f\left( x \right)} \right)^5} + f\left( x \right) + x = 0$?

You don't. You can't. It's not true. If there were a differentiable function satisfying that equation, then, differentiating both sides, $5f(x)^4f'(x)+ f'(x)= f'(x)(5f(x)^4+ 1)= 0$. So either f'(x)= 0, for all x, or 5f(x)^4+ 1= 0. If f'(x)= 0 for all x, then f(x) is a constant, c, such that $c^5+ c+ x= 0$ or $c^5+ c= -x$ which is impossible: the right side changes as x changes and the left side is a constant. However, all terms in 5f(x)^4+ 1 are non-negative: their sum is 0 only if 5f(x)^4= -1 which again says that f(x) is a constant.
• Feb 23rd 2009, 06:16 AM
FractalMath
Quote:

Originally Posted by HallsofIvy
You don't. You can't. It's not true. If there were a differentiable function satisfying that equation, then, differentiating both sides, $5f(x)^4f'(x)+ f'(x)= f'(x)(5f(x)^4+ 1)= 0$. So either f'(x)= 0, for all x, or 5f(x)^4+ 1= 0. If f'(x)= 0 for all x, then f(x) is a constant, c, such that $c^5+ c+ x= 0$ or $c^5+ c= -x$ which is impossible: the right side changes as x changes and the left side is a constant. However, all terms in 5f(x)^4+ 1 are non-negative: their sum is 0 only if 5f(x)^4= -1 which again says that f(x) is a constant.

What you say isn't true.

Differentiating both sides we get $5{\left( {f\left( x \right)} \right)^4}f'\left( x \right) + f'\left( x \right) + 1 = 0$. You're missing the +1.
• Feb 23rd 2009, 10:42 AM
HallsofIvy
Quote:

Originally Posted by FractalMath
What you say isn't true.

Differentiating both sides we get $5{\left( {f\left( x \right)} \right)^4}f'\left( x \right) + f'\left( x \right) + 1 = 0$. You're missing the +1.

Ouch! I need to have my eyes checked. Yes, you are completely correct (Actually it was the original "x" that has derivative one that I missed.)
So, what can you say about that equation?