Question

Examine if Rolle's theorem is applicable to any of the following functions. Can you say something about the converse of Rolle's theorem from these example ?

(i) $f\left(x\right)=\left[x\right]forxϵ[5,9]$

(i) $f\left(x\right)=\left[x\right]forxϵ[-2,2]$

(iii) $f\left(x\right)=x^2-1forxϵ[5,9]$.

Answer 1

(i) Given function is f(x)=f(x),x $ϵ[5,9]$

Since, f(x) is neither continuous nor derivable at integral point i.e., at 5,6,7,8 and 9.

Thus, f(x) is not continuous on [5,9].So, f(x) is not derivable on (5,9).

Using property, if f(x) is not continuous then it is not differentiable.

Moreover, f(5)=[5]=5 and f(9)=[9]=9 i.e., $f\left(5\right)\ne f\left(9\right)$.

Hence, Rolle's theorem is not applicable to f(x) in the [5,9].

It may be noted that f'(x)=0 for all non-integral points in [5,9].

This shows that converse of Rolle's theorem is not true.

(i) Given function is f(x)=f(x),x $ϵ[5,9]$

Since, f(x) is neither continuous nor differentable at integral point i.e., at -2.0,1,2

Thus, f(x) is not continuous on [-2,2].

So, f is not deirvable on (-2,2).

Moreover, $f(-2)=[-2]=-2\ne 2=f\left(2\right)$ f.e., $f\left(5\right)\ne f\left(9\right)$.

Hence, Rolle's theorem is not applicable to f(x) in the given interval.

It may be noted that f'(x)=0 for all non-integral points in [-2,2].

Given, f(x) = $x^2-1$, Which is a polynomial function. It is continuous and derivable at all $xϵR$.

In particular, f(x) is continuous on [1,2] and derivable on (1,2)

$f\left(1\right)=1^2-1=0$ and $f\left(2\right)=2^2-1=3$ i.e., $f\left(1\right)\ne f\left(2\right)$.

$\therefore$ Rolle's theorem is not applicable to given function in the given interval. Note that f'(x) = 2x for any x in (1,2)

Conclusion From the above examples, we conclude that the converse of Rolle's theorem does not hold. This means that it conditions fo Rolle's theorem doed theorem are not satified by a function f(x) on [a,b], then f'(x) may or may not vanish at some point in (a,b),