Determine where f is discontinuous. $f(x)=\left\{\begin{array}{ccc} \sqrt{-x} & \text { if } & x

Question

Determine where f is discontinuous. $f(x)=\left\{\begin{array}{ccc}
\sqrt{-x} & \text { if } & x<0 \
11-x & \text { if } & 0 \leq x<11 \
(11-x)^{2} & \text { if } & x>11
\end{array}\right.$
A) $0 \text { and } 11$
B) 0 only
C) $-11$ only
D) 11 only
E) $0 \text { and }-11$

Quizplus · Accepted Answer

1. The function is discontinuous at $x=0$ because the limit as $x$ approaches $0$ from the negative side is non-existent (approaches infinity) while the limit as $x$ approaches $0$ from the positive side is equal to $11$. Hence, the function jumps at $x=0$.
2. The function is discontinuous at $x=11$ because the limit as $x$ approaches $11$ from the negative side is equal to $0$ while the limit as $x$ approaches $11$ from the positive side is equal to $0$. However, the value of the function at $x=11$ is equal to $0$ for the linear part of the function and equal to $1$ for the squared part of the function. Hence, the function has a removable discontinuity at $x=11$ where we can redefine $f(11)$ as $0$ to make the function continuous.

Determine Where F Is Discontinuous A) 0 and 110 \text { and } 110 and 11

Determine Where F Is Discontinuous A) $0 \text { and } 11$