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 f(x)=x^2+ 14x +51 if x is less than -7
f(x)= 1 if x=-7
f(x) = −x^2−14x−47 if x is greater than -7

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

To show that the function f(x) has a removable discontinuity at x=-7, we need to examine the behavior of the function around that point.

First, let's check if f(x) is defined and continuous at x=-7. According to the given information, f(-7) is defined as 1. So f(x) is indeed defined at x=-7.

Next, let's check the continuity of f(x) at x=-7. To do this, we need to investigate the left and right limits as x approaches -7. Let's start by calculating the left limit:

lim(x->-7^-) [(x^2 + 14x + 51)] = lim(x->-7^-) [(x + 17)(x + 3)]

Since x + 17 and x + 3 are both polynomial functions, they are continuous everywhere, including x=-7. Therefore, we can substitute -7 into the expression:

lim(x->-7^-) [(x + 17)(x + 3)] = (-7 + 17)(-7 + 3) = 10(-4) = -40

Now let's calculate the right limit:

lim(x->-7^+) [(-x^2 - 14x - 47)] = lim(x->-7^+) [-(x + 17)(x + 3)]

Again, since x + 17 and x + 3 are continuous everywhere, including x=-7, we can substitute -7:

lim(x->-7^+) [-(x + 17)(x + 3)] = -(-7 + 17)(-7 + 3) = 10(-4) = -40

Both the left and right limits are equal to -40, but this value is not equal to f(-7) = 1. Therefore, f(x) is discontinuous at x=-7.

To make f(x) continuous at x=-7, we would need to redefine f(-7) to be equal to the value that the left and right limits converge to, which is -40 in this case.

So, we must redefine f(-7) as -40 to make f(x) continuous at x=-7.

Now, let's graph f(x):

Graphing f(x) requires plotting the different parts of the function separately.

For x less than -7, we have f(x) = x^2 + 14x + 51. This represents a parabola opening upwards.

For x greater than -7, we have f(x) = -x^2 - 14x - 47. This represents a parabola opening downwards.

Finally, for x = -7, we have f(x) = 1, which is just a single point on the graph.

Combining these pieces, we can draw the graph of f(x) to visualize the function.