What does this function show at x = 5? f(x)=x^2−25/x−5
A) removable discontinuity
B) jump discontinuity
C) infinite discontinuity
D) continuity
E) none of the above
at x=5, f(x) = 0/0, or undefined.
At any other value, though,
f(x) = (x-5)(x+5)/(x-5) = x+5
So, f(x) is undefined only at x=5. By defining f(5) = 10, the discontinuity is removed. f(x) = x+5 for all x.
So would it be none of the above?
Ummm. Do you not think it would be A?
I explained how the discontinuity could be removed. Hence, it is removable.
To determine what the function shows at x = 5, we need to evaluate the function at that point.
The given function is f(x) = (x^2 - 25) / (x - 5).
To evaluate the function at x = 5, substitute 5 into the function:
f(5) = (5^2 - 25)/(5 - 5)
Evaluating the numerator:
= (25 - 25) / (5 - 5)
= 0 / 0
At this point, we cannot divide by zero, which results in an indeterminate form.
By observing the function, we can see that (x - 5) is a factor in both the numerator and denominator. Therefore, we can simplify the function by factoring:
f(x) = (x + 5)(x - 5)/(x - 5)
Now, the (x - 5) terms cancel out:
f(x) = (x + 5), where x ≠ 5
We've simplified the function to f(x) = x + 5, with the condition that x ≠ 5.
This means that at x = 5, the function is undefined as a result of the division by zero.
Therefore, the correct answer is: A) removable discontinuity