Solving x^{101}+x^{83}+x=1 with Rolle's Theorem

[Yankel]

Hello, I need to find the solutions of this equation using Rolle's theorem

(the one saying that if f is a continuous and differentiated function on [a,b] then there exist a point c (a<c<b) such that f'(c)=0 )$$x^{101}+x^{83}+x=1$$

Thanks...

 

[Opalg]

Hello, I need to find the solutions of this equation using Rolle's theorem

(the one saying that if f is a continuous and differentiated function on [a,b] then there exist a point c (a<c<b) such that f'(c)=0 )$$x^{101}+x^{83}+x=1$$

Thanks...

You have left out a crucial condition of Rolle's theorem, namely the requirement that $f(a) = f(b)$.

Let $f(x) = x^{101}+x^{83}+x - 1$. Then $f'(x) = 101x^{100} + 83x^{82} + 1$, which is always positive, never zero. Rolle's theorem tells you that the equation $f(x)=0$ can have at most one solution. (If there were two points where $f(x)=0$ then Rolle's theorem would tell you that $f'(x)=0$ at some intermediate point.)

Can you see why the equation $f(x)=0$ must have one solution, and can you narrow down the interval in which that solution occurs? (You will not be able to find an exact solution, using Rolle's theorem or any other method.)

 

[Yankel]

You have left out a crucial condition of Rolle's theorem, namely the requirement that $f(a) = f(b)$.

Let $f(x) = x^{101}+x^{83}+x - 1$. Then $f'(x) = 101x^{100} + 83x^{82} + 1$, which is always positive, never zero. Rolle's theorem tells you that the equation $f(x)=0$ can have at most one solution. (If there were two points where $f(x)=0$ then Rolle's theorem would tell you that $f'(x)=0$ at some intermediate point.)

Can you see why the equation $f(x)=0$ must have one solution, and can you narrow down the interval in which that solution occurs? (You will not be able to find an exact solution, using Rolle's theorem or any other method.)

yes, I can see it now, thank you !