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question http://s30.postimg.org/6jnj4ie75/Untitled.png

I'm thinking that I need to do a proof by counter example? Is it possible to use Rolle's theorem:

f is cts on [a,b] and differentiable on (a,b) and f(a)=f(b), so there exists a c in (a,b) such that f'(c)=0.

Thanks

My answer:

Assume there exists c,d in (0,1) (c not equal to d) such that f'(c)=f'(d)=0.

$f'(x)=3x^2-3$

$f'(c)=3c^2-3$ so c=1 or -1. Contradiction.

$f'(d)=3c^2-3 $ so d=1 or -1. Contradiction.

user127700
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An alternative approach. In order to have 2 roots in $[0,1]$ it must contain a turning point within the vertical strip bounded by $ (x_0, x_1) = (0,1)$. The x value of the turning points are where the derivative is 0 $$ f'_m(x) = 3x^2 -3$$ $$ 3(x + 1)(x - 1) = 0$$

$x = -1$ is well outside the interval and is disqualified. $x = 1$ is on the border and is also disqualified , it has no chance to 'turn' and hit the x-axis again.

neofoxmulder
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