Problem

Find $x_{1}$ and $x_{2}$ where $f(x)$ is discontinuous (if they exist). Suppose $x_{1}

Solution

Step 1 :The function \(f(x)\) is defined for all real numbers. However, it is discontinuous at \(x=7\) because the function value at \(x=7\) is not equal to the limit of the function as \(x\) approaches 7. This is a removable discontinuity because the function is defined at \(x=7\) but the function value at \(x=7\) is not equal to the limit of the function as \(x\) approaches 7.

Step 2 :Let's calculate the limit of the function as \(x\) approaches 7 and the function value at \(x=7\).

Step 3 :\(x = 7\)

Step 4 :\(f = 4*x + 4\)

Step 5 :\(\text{limit}_f_{\text{at}_7} = 32\)

Step 6 :\(f_{\text{at}_7} = 22\)

Step 7 :The limit of the function as \(x\) approaches 7 is 32, but the function value at \(x=7\) is 22. Therefore, the function is discontinuous at \(x=7\). There are no other points of discontinuity.

Step 8 :\(\boxed{\text{Final Answer: The function is discontinuous at } x=7. \text{This is a removable discontinuity. There are no other points of discontinuity, so } x_{2} \text{ does not exist. Therefore, } x_{1}=7 \text{ and } x_{2}=\text{DNE}.}\)

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