Question

A function f is defined by $f\left(x\right)=2+{(x-1)}^{\frac{2}{3}}$ in $[0,2]$. Which of the following is not correct?

• A. $f$ is not derivable $(0,2)$
• B. $f$ is not continuous in $[0,2]$
• C. $f\left(0\right)=f\left(2\right)$
• D. Rolle's theorem is correct in$[0,2]$

Answer 1

Answer: (B), (D)

Rolle's theorem is correct in$[0,2]$

The explanation for the correct option:

Step 1: Finding the differentiability:

Given, the function, $f\left(x\right)=2+{(x-1)}^{\frac{2}{3}}$ in $\left[0,2\right]$

$$\begin{gathered} f\text{'}\left(x\right)=0+\frac{2}{3}\times \frac{1}{{\left(x-1\right)}^{{\displaystyle \frac{1}{3}}}} \\ =\frac{2}{3}\times \frac{1}{{\left(x-1\right)}^{{\displaystyle \frac{1}{3}}}} \end{gathered}$$

At $x=1$,

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

Thus, f(x) is not differentiable.

Step 2: Finding the continuity of the function:

At $x=1$,$f\left(1\right)=2+{(1-1)}^{\frac{2}{3}}$

$=2$

Thus, $f$ is continuous in $[0,2]$.

Step 3: Finding $f\left(0\right)$and $f\left(2\right)$:

$$\begin{gathered} f\left(0\right)=2+{(0-1)}^{\frac{2}{3}} \\ =2+1 \\ =3 \end{gathered}$$

$$\begin{gathered} f\left(2\right)=2+{(2-1)}^{\frac{2}{3}} \\ =2+1 \\ =3 \end{gathered}$$

Thus $f\left(0\right)=f\left(2\right)$

Step 4: Check the function to satisfy Rolle's theorem:

For the function to satisfy Rolle's theorem

It should satisfy the three conditions that is differentiability, continuity and $f\left(0\right)=f\left(2\right)$.

But it does not satisfies all the three condition.

Thus, the function does not satisfy Rolle's theorem at $[0,2]$.

Hence option (C),(D) is the correct answer.