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

$ f(x) = \left\{ \begin{array}{ll} \dfrac{2x^2 - 5x - 3}{x - 3} & \mbox{if $ x \neq 3 $} \hspace{30mm} a = 3\\ 6 & \mbox{if $ x = 3 $} \end{array} \right.$

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Missouri State University

Harvey Mudd College

University of Nottingham

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This is problem number twenty two of the Stuart Calculus eighth edition Section two point fine. Explain what the function is discontinuous at the given number, eh? Sketch. The graph of the function affects is the peaceful dysfunction to X squared minus five. Expand its three about brother quantity X minus three if X is not equal to three and the functions equals six if X is equal to three and our day is three. So we will approach this by presenting the equation for continuity this definition and Sena belts too for a equals two three when we first attempt the limit as experts. Just three of the function to X squared minus five explain. It's three divided by X minus three No, we can make a simplification and factor the integrator into two x plus one and next minus three. And we do it this way because two extremes Exes to X squared to accepting the freezing of six plus one is native fanatics. And then one times to think of three is equal to the other three and the denominator. We also have a negative three ex ministry and these two terms cancel. And that leaves us with limited as expert history of dysfunction two X plus one, which gives us two times three plus one, which is equal to seven. And unfortunately this is not equal. The function evaluated at three, which we are given A six, has shown here the function. So because the limit is not equal to the after they we know that it's not continuous. And if we were to plot this on the graph here, I will see that Ah, the function is to find that three, six, one, two, three, four, five, six. However, this function, um who's from is it is a linear form. Dysfunction to X squared minus five Ex ministry over X minus three Uh, it has a removal discount at X equals three. Otherwise he would equal seven x equals three. And if we want this linear function, we see that it has this linear shape again. Dysfunction is linear because has a removable discard nodia three, which is what? Which explains the whole here. This point is continuity. Paddocks equals three on DH. That results because of ex ministry, and the denominator from the function is to find it. X equals three y equal six and even though the limited exists and is equal to seven, the function is not defined there and therefore we say that it is discontinuous because of this point, this continuity.

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