Answer to Question #103931 in Calculus for BIVEK SAH

Question

Examine the continuity of the function f:[1,4]→R defined by f(x)= [ x]+4/(3x-1)

Answer 1

Note: we assume that $y=[x]$ means $y=\lfloor x\rfloor$. Which in turn means

$$\lfloor x\rfloor=\left\{\left.m\in\mathbb{x}\right|m\le x\right\}$$

( More information: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions )

Since, $f : [1;4]\to\mathbb{R}$ , then consider how the specified function $y=\lfloor x\rfloor$ behaves on a given interval:

$$\lfloor x\rfloor=\left\{\begin{array}{c} 1,\,\,\,\text{if}\,\,\,1\le x<2\\ 2,\,\,\,\text{if}\,\,\,2\le x<3\\ 3,\,\,\,\text{if}\,\,\,3\le x<4\\ 4,\,\,\,\text{if}\,\,\,x=4 \end{array}\right.$$

Then, the function $f(x)$ looks like this

$$f(x)=\left\{\begin{array}{l} \displaystyle\frac{5}{3x-1},\,\,\,\text{if}\,\,\,1\le x<2\\[0.5cm] \displaystyle\frac{6}{3x-1},\,\,\,\text{if}\,\,\,2\le x<3\\[0.5cm] \displaystyle\frac{7}{3x-1},\,\,\,\text{if}\,\,\,3\le x<4\\[0.5cm] \displaystyle\frac{8}{3x-1},\,\,\,\text{if}\,\,\,x=4 \end{array}\right.$$

We see that it is necessary to study the continuity of the function at the points $x=2$ , $x=3$ ,

$x=4$. For this, one-sided limits must be considered.

( More information: https://en.wikipedia.org/wiki/One-sided_limit )

In our case,

$$\lim\limits_{x\to 1^+}f(x)=\lim\limits_{x\to 1^+}\frac{5}{3x-1}=\frac{5}{3\cdot 1-1}=\frac{5}{2}\\[0.5cm] f(1)=\frac{5}{3\cdot 1-1}=\frac{5}{2}\\[0.5cm] \boxed{\lim\limits_{x\to 1^+}f(x)=\frac{5}{2}=\frac{5}{2}=f(1)}\\[0.5cm] \lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}\frac{5}{3x-1}=\frac{5}{3\cdot 2-1}=1\\[0.5cm] \lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}\frac{6}{3x-1}=\frac{6}{3\cdot 2-1}=\frac{6}{5}\\[0.5cm] \boxed{\lim\limits_{x\to 2^-}f(x)=1\neq\frac{6}{5}=\lim\limits_{x\to 2^+}f(x)}\\[0.5cm] \lim\limits_{x\to 3^-}f(x)=\lim\limits_{x\to 3^-}\frac{6}{3x-1}=\frac{5}{3\cdot 3-1}=\frac{6}{8}\\[0.5cm] \lim\limits_{x\to 3^+}f(x)=\lim\limits_{x\to 3^+}\frac{7}{3x-1}=\frac{7}{3\cdot 2-1}=\frac{7}{8}\\[0.5cm] \boxed{\lim\limits_{x\to 3^-}f(x)=\frac{6}{8}\neq\frac{7}{8}=\lim\limits_{x\to 3^+}f(x)}\\[0.5cm] \lim\limits_{x\to 4^-}f(x)=\lim\limits_{x\to 4^-}\frac{7}{3x-1}=\frac{7}{3\cdot 4-1}=\frac{7}{11}\\[0.5cm] f(4)=\frac{8}{3\cdot4-1}=\frac{8}{11}\\[0.5cm] \boxed{\lim\limits_{x\to 4^-}f(x)=\frac{7}{11}\neq\frac{8}{11}=f(x)}\\[0.5cm]$$

Conclusion,

The function $f(x)$ is discontinuous at the points $x=2$ , $x=3$ , $x=4$ .

The function $f(x)$ is continuous at the point $x=1$ .