At what points are the functions in Exercises 13–30 continuous...

Question

At what points are the functions in Exercises 13–30 continuous?

g(x) = { (x² − x – 6)/(x – 3), x ≠ 3

5, x = 3

Accepted Answer

Welcome back, everyone. Determine the points where the function below is continuous. Y is a piecewise function such that it is equal to x2 minus 9 divided by X minus 3 if X is not equal to 3 and 6 if X is equal to 3. So essentially we can analyze each part of the piece wise function. First of all, we can see that we have X2 minus 9. Divided by x minus 3 and a potential or actually a point of discontinuity in this function is x equals 3 because division by 0 is undefined, right? And for the second part we have 6, which is defined for all real numbers. Because it is a constant, right? So the only point of this continuity that we can have is X equals 3, and essentially what we have to do is analyze this point using The definition of a continuous function. We want to make sure that the limit as x approaches the point of discontinuity, in this case 3 of F of X, let's call our function F of X is equal to the value of the function at that point. So F of 3, right? So, we know that F of 3 is 6. And now we want to analyze the limits. So for the limit, we're going to take the first part of the function, right? So let's evaluate the limit as x approaches 3 of X2 minus 9 divided by X minus 3. If we simply attempt the direct substitution, this gives us 32 minus 9 divided by 3 minus 3, and this is an indeterminate form which is 0 divided by 0. So we're going to write down the indeterminant form. And then we can notice that we can factor it out to eliminate a whole. So what we're going to do is apply the difference of squares factorization, and we can factor out X2 minus 3 squad as X minus 3 multiplied by X + 3. Which is then divided by X minus 3, and this eliminates the whole. We can cancel out the factor X minus 3. And once again attempt direct substitution which gives us 3 + 3. This gives us 6. So what have we shown? Well, we have shown that the limit as X approaches 3. Of F of X. is equal to 6, which is also equal to the value of the function at X equals 3. So the function is actually continuous at 3 as well. So the only potential point of discontinuity was proven to be a function where It was proven to be a point where the function is continuous, right? Meaning the function is continuous for all real numbers. We can say that F of X is continuous. Or all acts. Belonging to real numbers, and that will be our final answer for this problem. Thank you for watching.

At what points are the functions in Exercises 13–30 continuous?

g(x) = { (x² − x – 6)/(x – 3), x ≠ 3
5, x = 3

Key Concepts

Continuity of Functions

Limits

Piecewise Functions

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