A function f(x) is said to have a removable discontinuity at x=a if:

1. f is either not defined or not continuous at x=a.
2. f(a) could either be defined or redefined so that the new function IS continuous at x=a.

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Let
Show that f(x) has a removable discontinuity at x=−7 and determine what value for f(−7) would make f(x) continuous at x=−7.
Must redefine f(−7)=_____________.
Now for fun, try to graph f(x). It's just a couple of parabolas!

f(x) is not given.

As an example, if f(x) is defined as follows:
f(x)=x² for x<0, and
f(x)=2x² for x>0.
Graph f(x) and you will find x=0 is undefined.
Since Lim f(x) x->0- equals Lim f(x) x->0+, we say that there is a removable discontinuity at x=0. The discontinuity can be removed by redefining f(x).

I'm sorry, I was having some problems posting the questions....

Again
Let

f(x)= mx-12 if x is less than -5
x^2 +5x - 7 if x is greater than -5
Show that f(x) has a removable discontinuity at x=−7 and determine what value for f(−7) would make f(x) continuous at x=−7.
Must redefine f(−7)=_____________.

OK I made a mistake... AGAIN fx is not equal to that sorry. I will repost this question.

To show that f(x) has a removable discontinuity at x = -7, we need to check if f(x) is either not defined or not continuous at x = -7.

First, let's consider the definition of f(x) given in the problem. It is not explicitly stated, so we would need additional information or an equation for f(x) to proceed.

Assuming we have an equation for f(x), we can check if it is defined and continuous at x = -7.

To determine if f(x) is defined at x = -7, we need to check if there is a value assigned for f(-7) in the equation.

To determine if f(x) is continuous at x = -7, we need to check if the limit of f(x) as x approaches -7 exists and equals f(-7).

Assuming f(x) is defined as a piecewise function, let's consider two parabolas for simplicity.

Let the equation of the first parabola be f(x) = x^2 - 1 for x ≤ -7.

Let the equation of the second parabola be f(x) = x^2 + 3 for x > -7.

To check if f(x) is defined at x = -7, we substitute x = -7 into both equations.

For the first parabola, f(-7) = (-7)^2 - 1 = 49 - 1 = 48.

For the second parabola, f(-7) is not defined since the equation is only valid for x > -7.

Therefore, f(x) is not defined at x = -7 according to the second condition for a removable discontinuity.

Now, to make f(x) continuous at x = -7, we need to redefine f(-7) such that it fills the gap between the two parabolas.

Since we want f(x) to be continuous, we can redefine f(-7) as the limit of f(x) as x approaches -7.

Taking the limit from the left, lim(x→-7-) (x^2 - 1) = (-7)^2 - 1 = 49 - 1 = 48.

Therefore, we redefine f(-7) = 48 to make f(x) continuous at x = -7.

Now, let's try to graph f(x).

The graph of f(x) would consist of two separate parabolas, one defined for x ≤ -7 and the other for x > -7. The first parabola would have the equation y = x^2 - 1 and the second parabola would have the equation y = x^2 + 3.

For x ≤ -7, the graph would be a downward opening parabola shifted vertically downward by 1 unit compared to the standard parabola y = x^2.

For x > -7, the graph would be an upward opening parabola shifted vertically upward by 3 units compared to the standard parabola y = x^2.

Joining the two parts of the graph with an open circle at the point (-7, 48) would represent the removable discontinuity.