Question

For which, interval, the function$\frac{x^2-3x}{x-1}$ satisfies all the conditions of Rolle’s theorem

• A. $[0,3]$
• B. $[-3,0]$
• C. $[1.5,3]$
• D. For no interval

Answer 1

The correct option is D

For no interval

Explanation for all the options:

Let$f\left(x\right)$ is a function then it will follow Rolle's theorem if,

$1.f\left(x\right)$ is a continuous function for the interval$[a,b]$.

$2.f\left(x\right)$ is a differentiable function for the interval$(a,b)$.

$3.f\left(a\right)=f\left(b\right)$ the function has an equal value at endpoints.

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

let's check the value of the functions at the endpoints of all the intervals,

$$\begin{gathered} \Rightarrow f\left(0\right)=\frac{0(0-3)}{0-1}=0 \\ \Rightarrow f\left(3\right)=\frac{3(3-3)}{3-1}=0 \\ \Rightarrow f(-3)=\frac{-3(-3-3)}{-3-1}=\frac{-18}{-4}=\frac{9}{2} \\ \Rightarrow f(1.5)=\frac{{\displaystyle \frac{3}{2}}(\frac{3}{2}-3)}{\frac{3}{2}-1}=\frac{{\displaystyle \raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\frac{9}{2} \end{gathered}$$

clearly, just two values may follow Rolle's theorem if$f\left(x\right)$ follows the other two remaining conditions,

$f\left(0\right)=f\left(3\right)=0$

Explanation for option$\left(A\right)$:

$[0,3]$ just one interval, for which$f\left(x\right)$ will follow Rolle's theorem as$f\left(0\right)=f\left(3\right)=0$.

but$f\left(x\right)$ is not a continuous function$f\left(x\right)$ at$x=1$, it means$f\left(x\right)$ is not a continuous function in interval$[0,3]$.

therefore, $f\left(x\right)$ does not satisfies all the conditions of Rolle’s theorem for the interval$[0,3]$.

Explanation for option$\left(B\right)$:

We can see from the above$f(-3)\ne f\left(0\right)$.

so, $f\left(x\right)$ does not satisfies all the conditions of Rolle’s theorem for the interval$[-3,0]$.

Explanation for option$\left(C\right)$:

We can see from the above$f(1.5)\ne f\left(3\right)$.

so, $f\left(x\right)$ does not satisfies all the conditions of Rolle’s theorem for the interval$[-3,0]$.

Explanation for option$\left(D\right)$:

therefore, $f\left(x\right)$ does not satisfies all the conditions of Rolle’s theorem for no interval.

Hence, option$\left(D\right)$ is the correct option.