Question

Consider the function $f\left(x\right)=\{\begin{array}{ccc}\sqrt{-x}& if& x<0\\ 0& if& 0\le x\le 4\\ x-4& if& x>4\end{array}$. Choose the answer which best describes the continuity of this function:

• A. The function is unbounded and therefore cannot be continuous.
• B. The function is right continuous at $x=0$.
• C. The function has a removable discontinuity at 0 and 4, but is continuous on the rest of the real number line.
• D. The function is continuous on the entire real number line.

Answer 1

Answer: (D)

The function is continuous on the entire real number line.

$f\left(x\right)=\left[\begin{array}{ccc}\sqrt{-x}& if& x<0\\ 0& if& 0\le x\le 4\\ x-4& if& x>4\end{array}\right]$
Checking for continuity at $x=0$:
${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\sqrt{x}=0{lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}0=0f\left(0\right)=0$.
So the function is continuous at $x=0$.
Checking for continuity at $x=4$:
${lim}_{x\to 4^{-}}f\left(x\right)={lim}_{x\to 4^{-}}0=0{lim}_{x\to 4^{+}}f\left(x\right)={lim}_{x\to 4^{+}}(x-4)=4-4=0f\left(4\right)=0$.
So the function is continuous at $x=4$.
Since this is an algebraic function continuous at both the points where its definition changes, it is continuous everywhere on the real number line. Hence, option D is the correct answer.