# Plz solve part 4 of question 4

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

Differentiating with respect to x, we get

$f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{2}{3}-1}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{2}{3}{\left(x-2\right)}^{\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}⇒f\text{'}\left(x\right)=\frac{2}{3{\left(x-2\right)}^{\frac{1}{3}}}$

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

Therefore, $f\left(x\right)$ is not derivable on .

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 is given to be .

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

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

Now, let $c=0$.

Then,

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

Thus, is discontinuous at = 0.

Therefore, $f\left(x\right)$ is not continuous in .

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

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

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

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

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

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

But

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 .
The domain of is given to be .

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

At x = 0, we have

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

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

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

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