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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 $} \end{array} \right.$

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given the piecewise function F of x which is equal to two, X squared minus five X minus three over x minus three. If x is not equal to three and f of x equals six. If X equals three, we need to know why the function is not continuous at A equal to three to do this, we begin by checking conditions for continuity and the first one would be is F defined at x equals three. Now, since actually goes three is part of the domain of this function, then we have F of three. This is equal to six and this is defined. I actually want to verify if the limit of this function As X approaches three exists And since the limit as X approaches three of F of X, this is equal to The limit as X approaches three of two, x squared -5, X -3 Over X -3 which simplifies to The limit as x approaches three of two, X plus one times x minus three over x minus three, Which cancels out the X -3. And gives us the limit as X approaches three of two, X plus one and evaluating gives us two times three plus one equals seven. We say that the limit exists and unless you want to check if The limit of the function equals the value of the function at three. Now, since the limit of F of X As X approaches three equals 7, Which is not equal to six, which is the value of the function at three, then you say that this condition does not hold. And So the function is discontinuous at three. And for the graph of the function we will use the information we had from the previous steps. So in here we have f. F three, which is equal to six. The limit of the function as x approaches three equal 7. And previously have shown that f of X can be reduced to X plus one, Where X is not equal to three. So the first thing we have to do is to plot the point F of three equal six, which will be over here. Next we want a plot the function two, X plus one where X is not equal to three and we not hear that the function has a whole at X equals three. Because when we reduce this function, We were able to cancel out X -3. Now to find the y coordinate of this whole, we simply plug in X equals 3, 2. The reduced form of the function. So it will be what is going to be two times three plus one. That's equal to seven. So the whole has coordinates 3, 7. And we can put it here, which would be this point. And then to plot F of X which is equal to two, X plus one. We simply pick X values and we solve for F for the Y coordinate. So let's say X is equal to two and we have FF two, that's two times 2 plus one That's equal to five. And if XAT equals four, then we have fo for which is just two times four plus one, or that's nine. And so we plot 25 which will be this point, And then we plot 4 9, which will be this point. And then from here we connect the dots and we have the grab for two X plus one. And so this is the graph of our function.

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