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.

Let f(x)= x2+10x+26 2 −x2−10x−24 if x−5 if x=−5 if x−5
Show that f(x) has a removable discontinuity at x=−5 and determine what value for f(−5) would make f(x) continuous at x=−5.
question:
Must redefine f(−5)=

A function with a removable discontinuity could be redefined to remove the discontinuity if, at the point of removable discontinuity (x=-5) in this case, Lim f(x) x->-5- and x->-5+ are equal.

The discontinuity can be removed by redefining f(x) such that f(-5)=one of the above limits.

As I am unable to read unambiguously the definition of the function f(x), I am not able to show that f(-5) should be.

To show that f(x) has a removable discontinuity at x = -5, we need to determine if f(x) is either not defined or not continuous at x = -5.

First, let's consider the functions in the given expression:
f1(x) = x^2 + 10x + 26
f2(x) = 2 - x^2 - 10x - 24 if x ≠ -5
f2(x) = x - 5 if x = -5

Now, let's evaluate f(x) at x = -5 using the given expression:
f(-5) = f1(-5) * f2(-5)
= (-5)^2 + 10(-5) + 26 * (-5 - 5)
= 25 - 50 + 26 * (-10)
= 1 + 26 * (-10)
= 1 - 260
= -259

So, f(-5) is equal to -259. This means that f(x) is defined at x = -5.

Next, let's consider the limit of f(x) as x approaches -5. We'll evaluate this limit from both sides:

As x approaches -5 from the left (x < -5):
lim(x->-5-) f(x) = lim(x->-5-) (x^2 + 10x + 26) * f2(x)
= lim(x->-5-) (x^2 + 10x + 26) * (x - 5)
= (-5^2 + 10(-5) + 26) * (-5 - 5)
= (25 - 50 + 26) * (-10)
= 1 * (-10)
= -10

As x approaches -5 from the right (x > -5):
lim(x->-5+) f(x) = lim(x->-5+) (x^2 + 10x + 26) * f2(x)
= lim(x->-5+) (x^2 + 10x + 26) * (x - 5)
= (-5^2 + 10(-5) + 26) * (-5 - 5)
= (25 - 50 + 26) * (-10)
= 1 * (-10)
= -10

Since the limit from both sides is the same, we can say that lim(x->-5) f(x) exists.

Now, to make f(x) continuous at x = -5, we need to redefine f(-5) to be equal to the value of the limit.

Therefore, we must redefine f(-5) = -10 in order to make f(x) continuous at x = -5.

To determine if function f(x) has a removable discontinuity at x=-5, we need to verify the conditions mentioned above:

1. Check if f(x) is either not defined or not continuous at x=-5:
- The function f(x) is defined for all values of x, including x=-5. So it is defined at x=-5.
- To check if f(x) is continuous at x=-5, we need to evaluate the left-hand limit (lim f(x)) as x approaches -5 and compare it with the right-hand limit (lim f(x)) as x approaches -5.

2. Evaluate the left-hand limit:
- For values of x less than -5, f(x) is given by f(x) = (x^2 + 10x + 26) / (2 - x^2 - 10x - 24) = (x^2 + 10x + 26) / (-x^2 - 10x - 22)
- Taking the limit as x approaches -5 from the left-hand side: lim f(x) = lim [(x^2 + 10x + 26) / (-x^2 - 10x - 22)] as x approaches -5
- By direct substitution, we get lim f(x) = (-5^2 + 10*(-5) + 26) / (-(-5^2) - 10*(-5) - 22) = (25 - 50 + 26) / (25 - 50 - 22) = 1 / 3

3. Evaluate the right-hand limit:
- For values of x greater than -5, f(x) is given by f(x) = x - 5
- Taking the limit as x approaches -5 from the right-hand side: lim f(x) = lim (x - 5) as x approaches -5
- By direct substitution, we get lim f(x) = (-5 - 5) = -10

4. Compare the left-hand and right-hand limits:
- Since the left-hand limit (1/3) is not equal to the right-hand limit (-10), f(x) is not continuous at x=-5.

To make f(x) continuous at x=-5, we need to redefine f(-5) to match the left-hand and right-hand limits.
We already have lim f(x) = 1/3, so we must redefine f(-5) = 1/3 to remove the discontinuity and make f(x) continuous at x=-5.