Proving at Most One Real Root of x³-15x+C in [-2,2]

[alexk307]

Homework Statement

Show that the equation x³-15x+C=0 has at most one real root in [-2,2]

Homework Equations

Rolle's Theorem, and Intermediate Value Theorem.

The Attempt at a Solution

I showed that there is a root in [-2,2] by use of Intermediate Value Theorem.

f(-2)<0
f(2)>0


But then to show there is not two roots, I tried to use Rolle's theorem which says that if f'(x)=0 then there must be two points f(a)=f(b). I found that f'(x)=0 at sqrt(5). Which I then tried to plug back in the original function in hopes that this point would lie above the x axis, therefore it wouldn't cross the x-axis twice. But because of the C I cannot prove this because I can always make C smaller and smaller in order to make f(sqrt(5))<0.

 

[micromass]

Rolle's theorem tells us that any function which is continuous on [a,b], differentiable on ]a,b[ and has the property that f(a)=f(b), must have a c such that $$f'(c)=0$$.

Now, assume that f(x)=x³-15x+C has two roots a and b in [-2,2]. Then f(a)=f(b). So what does Rolle's theorem yield?

 

[alexk307]

Rolle's theorem tells us that any function which is continuous on [a,b], differentiable on ]a,b[ and has the property that f(a)=f(b), must have a c such that $$f'(c)=0$$.

Now, assume that f(x)=x³-15x+C has two roots a and b in [-2,2]. Then f(a)=f(b). So what does Rolle's theorem yield?

oh! okay so that sqrt(5) is beyond the interval therefore proving that there is only one root in [-2,2].
Thank you

 

[micromass]

correct!