solution of equation/ rolle's theorem

**serious331** · May 2nd 2010, 01:27 PM

For c > 0, prove that the following equation does not have two solutions:

$x^3 - 3x + c = 0$ $; 0 < x < 1$
I think we have to show this using Rolle's theorem, but I am not being able to use the theorem properly on this question. Can anyone give any suggestions?

**maddas** · May 2nd 2010, 01:35 PM

The LHS is decreasing (differentiate). Can a decreasing function have two zeros?

**serious331** · May 2nd 2010, 01:39 PM

Originally Posted by maddas

The LHS is decreasing (differentiate). Can a decreasing function have two zeros?

hi maddas,

How is the function decreasing? Its derivative is positive, doesn't it mean that it is strictly increasing?

**maddas** · May 2nd 2010, 01:41 PM

Well, if its increasing, it can't have two zeros either. Is the derivative really positive?

edit: no, I'm pretty sure its negative.

**serious331** · May 2nd 2010, 02:04 PM

Originally Posted by maddas

Well, if its increasing, it can't have two zeros either. Is the derivative really positive?

edit: no, I'm pretty sure its negative.

Thanks maddas, it is negative...

so differentiating gives $3x^2 - 3 = 0$ . But then, $f'(x) = 0$ for $x= \pm 1$ .

**maddas** · May 2nd 2010, 02:09 PM

Fortunately you're only considering 0<x<1.

**serious331** · May 2nd 2010, 02:17 PM

Originally Posted by maddas

Fortunately you're only considering 0<x<1.

So , I would do the same for $x^5 + 5x + 1 = 0 ; -1<x<0$ to show that the equation has only one solution..?

**maddas** · May 2nd 2010, 02:19 PM

Pretty sure.

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