Problem

Explain why the function is discontinuous at the …

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Problem 19 Easy Difficulty

Explain why the function is discontinuous at the given number $ a $. Sketch the graph of the function.

$ f(x) = \left\{
\begin{array}{ll}
x + 3 & \mbox{if $ x \le -1 $} \hspace{40mm} a = -1\\
2^x & \mbox{if $ x > -1 $}
\end{array} \right.$

Answer

since the left-hand and the right-hand limits of $f$ at -1 are not equal, $\lim _{x \rightarrow-1} f(x)$ does not exist, and $f$ is discontinuous at -1

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Video Transcript

here, we have a piecewise defined function F of x F of x equals X plus three. When X is less than or equal to negative one, that's the red portion of the graph. And then for X greater than negative one F of X is defined to be uh to T D X. That's the blue portion of the grave. Uh This function is clearly uh Discontinuous when x equals -1. Uh to limit of the function as X approaches negative one looks to be about one half. Whereas to limit, let me rephrase that as X approaches negative one from the right side from the positive side, the function seems to be approaching a value close to about a half, whereas if we approach X equals negative one, coming in from the left side, uh to function is approaching a value of two. So the function is discontinuous because the limit of the function as X approaches uh negative one does not exist because the right hand limit as we approach negative one from the right side, uh is not going to be the same as the left hand limit as we approach negative one from the left side, from the negative side. Okay, so uh if a function uh does not have a limit because the right hand limit and the left hand limits are different, this is one half, this is too uh if the right hand limits and the left hand limits are different, the limit does not exist and a function cannot be continuous at a point. If the limit does not exist

Leon D.

Temple University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 5

Continuity

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