Use Rolle's theorem to show that

$f(x)=x^3-\frac{3}{2}x^2+\lambda$, $\lambda \in \mathbf{R}$

never has 2 zeroes in $[0,1]$.

I started by assuming that $\exists$ $2$ zeroes in$[0,1]$


$\exists a,b \in [0,1]$

$\ni f(a)=f(b)=0$

By Rolle's theorem,

$\exists f'(c)=0 for c \in ]a,b[$



When c=3, I get a contradiction since $c \in [0,1]$

But when $c=0$, does this mean that there is a stationary point at 0 or there is a point of inflexion at 0.


You have committed a small mistake in your calculations in this part:



Actually it should be like this:

$$f'(c)=0$$ $$3c^2-3c=0$$ $$3c(c-1)=0$$ $$c=0,1$$

But already it has been stated that $c \in (0,1)$ for Rolle's Theorem.

So any such $c$ does not exist and hence the initial hypothesis is false.

Thus, the equation can never have two roots in the interval [$0,1$].

  • $\begingroup$ How come, $[o,1]$ is a closed interval. It means $0$ and $1$ are included? $\endgroup$ – Tosh May 24 '16 at 12:42
  • $\begingroup$ [$0,1$] is a closed interval, it means $0$ and $1$ are included. But ($0,1$) is an open interval, it means $0$ and $1$ are excluded. $\endgroup$ – SchrodingersCat May 24 '16 at 12:46
  • $\begingroup$ the interval is a closed one as per the question. $\endgroup$ – Tosh May 24 '16 at 13:05
  • 1
    $\begingroup$ @SirusBlack Of course. Check that irrespective of whether the interval is closed or open, the $c$ in the Rolle's theorem always belongs to the open interval. Always. $\endgroup$ – SchrodingersCat May 24 '16 at 13:08
  • $\begingroup$ Thanks a lot. This is really helpful. $\endgroup$ – Tosh May 24 '16 at 13:22

$f'(x) = 3x(x-1) < 0$ for $x\in(0,1)$, hence $f(x)$ is decreasing on $[0,1]$ and $f(x)=-\lambda$ cannot have more than one solution in $[0,1]$.


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