Question

Discuss the applicability of Rolle's theorem for the following functions on the indicated intervals
(i) f(x) = 3 + (x − 2)^2/3 on [1, 3]

(ii) f(x) = [x] for −1 ≤ x ≤ 1, where [x] denotes the greatest integer not exceeding x

(iii) f(x) = sin$\frac{1}{x}$for −1 ≤ x ≤ 1

(iv) f(x) = 2x² − 5x + 3 on [1, 3]

(v) f(x) = x^2/3 on [−1, 1]

(vi) $f\left(x\right)=\left\{\begin{array}{ll}-4x+5,& 0\le x\le 1\\ 2x-3,& 1<x\le 2\end{array}\right.$

Answer 1

(i) The given function is $f\left(x\right)=3+{\left(x-2\right)}^{\frac{2}{3}}$.

Differentiating with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{2}{3}-1} \\ \Rightarrow f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{-1}{3}} \\ \Rightarrow f\text{'}\left(x\right)=\frac{2}{3{\left(x-2\right)}^{\frac{1}{3}}} \end{gathered}$$

Clearly, we observe that for x=2$\in \left[1,3\right]$, $f\text{'}\left(x\right)$ does not exist.

Therefore, $f\left(x\right)$ is not derivable on $\left[1,3\right]$.

Hence, Rolle's theorem is not applicable for the given function.

(ii) The given function is $f\left(x\right)=\left[x\right]$.
The domain of f is given to be $\left[-1,1\right]$.

Let $c\in \left[-1,1\right]$ such that c is not an integer.
Then,
$\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$

Thus, $f\left(x\right)$ is continuous at $x=c$.

Now, let $c=0$.

Then,

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=-1\ne 0=f\left(0\right)$

Thus, f is discontinuous at x = 0.

Therefore, $f\left(x\right)$ is not continuous in $\left[-1,1\right]$.

Rolle's theorem is not applicable for the given function.

(iii) The given function is $f\left(x\right)=\sin \frac{1}{x}$.
The domain of f is given to be $\left[-1,1\right]$.

It is known that $\underset{x\to 0}{\lim }\sin \frac{1}{x}$ does not exist.

Thus, $f\left(x\right)$ is discontinuous at x = 0 on $\left[-1,1\right]$.

Hence, Rolle's theorem is not applicable for the given function.


(iv) The given function is $f\left(x\right)=2x^2-5x+3$ on $\left[1,3\right]$.
The domain of f is given to be $\left[1,3\right]$.
It is a polynomial function.
Thus, it is everywhere derivable and hence continuous.

But
$$\begin{gathered} f\left(1\right)=0\mathrm{and}f\left(3\right)=6 \\ \Rightarrow f\left(3\right)\ne f\left(1\right) \end{gathered}$$

Hence, Rolle's theorem is not applicable for the given function.

(v) The given function is $f\left(x\right)=x^{\frac{2}{3}}$ on $\left[-1,1\right]$.
The domain of f is given to be $\left[-1,1\right]$.

Differentiating $f\left(x\right)$ with respect to x, we get

$f\text{'}\left(x\right)=\frac{2}{3}{x}^{-\frac{1}{3}}$
We observe that at $x=0$, $f\text{'}\left(x\right)$ is not defined.
Hence, Rolle's theorem is not applicable for the given function.

(vi) The given function is
$f\left(x\right)=\left\{\begin{array}{l}-4x+5,0\le x\le 1\\ 2x-3,1<x\le 2\end{array}\right.$

At x = 0, we have

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left[-4\left(1-h\right)+5\right]=1$
And
$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }\left[2\left(1+h\right)-3\right]=-1$

$\therefore$ $\underset{x\to 1^{-}}{\lim }f\left(x\right)\ne \underset{x\to 1^{+}}{\lim }f\left(x\right)$

Thus, $f\left(x\right)$ is discontinuous at $x=1$.
Hence, Rolle's theorem is not applicable for the given function.