Question

6. Given \( f(x)=\left\{\begin{array}{cl}x+5 & \text { if } x \leq-2 \\ x^{2} & \text { if }-21\end{array}\right. \) A) Find the numbers at which \( \mathrm{f} \) is discontinuous. Explain why \( \mat

Answer 1

Explanation:

To find the numbers at which the function ff is discontinuous, we need to identify any values of xx where there is a change in the definition of the function, leading to a jump or break in the graph of f(x)f(x).

Looking at the given function f(x)f(x), we see that it has two different definitions for different intervals of xx:

f(x)=x+5f(x)=x+5 for x≤−2x≤−2

f(x)=x2f(x)=x2 for x>−2x>−2

Now, let's identify the numbers where f(x)f(x) may be discontinuous:

x=−2x=−2: At x=−2x=−2, there is a change in the definition of f(x)f(x) from x+5x+5 to x2x2. This is a potential point of discontinuity because there's a jump in the function at this point.

x=−1x=−1: There is another potential point of discontinuity at x=−1x=−1, which is the boundary between the two intervals where the function is defined differently.

To determine whether these points are indeed discontinuities, we need to check if the limit of the function as xx approaches these points from both the left and the right is different.

As xx approaches -2 from the left (x<−2x<−2), the function is f(x)=x+5f(x)=x+5, so the limit from the left is:

lim⁡x→−2−f(x)=lim⁡x→−2−(x+5)=(−2)+5=3limx→−2−​f(x)=limx→−2−​(x+5)=(−2)+5=3

As xx approaches -2 from the right (x>−2x>−2), the function is f(x)=x2f(x)=x2, so the limit from the right is:

lim⁡x→−2+f(x)=lim⁡x→−2+(x2)=(−2)2=4limx→−2+​f(x)=limx→−2+​(x2)=(−2)2=4

Since the left-hand limit and the right-hand limit are not equal at x=−2x=−2, this is indeed a point of discontinuity.

As xx approaches -1 from the left (x<−1x<−1), the function is f(x)=x+5f(x)=x+5, so the limit from the left is:

lim⁡x→−1−f(x)=lim⁡x→−1−(x+5)=(−1)+5=4limx→−1−​f(x)=limx→−1−​(x+5)=(−1)+5=4

As xx approaches -1 from the right (x>−1x>−1), the function is f(x)=x2f(x)=x2, so the limit from the right is:

lim⁡x→−1+f(x)=lim⁡x→−1+(x2)=(−1)2=1limx→−1+​f(x)=limx→−1+​(x2)=(−1)2=1

Since the left-hand limit and the right-hand limit are not equal at x=−1x=−1, this is also a point of discontinuity.

Therefore, the numbers at which f(x)f(x) is discontinuous are x=−2x=−2 and x=−1x=−1, and these points are discontinuous because the limits from the left and the right do not match at these points.