Which of the following must be true for the graph of the function
f(x)= (x^2-25)/5x-25)
There is:
I. a removable discontinuity at x=5
II. a vertical asymptote at x=5
III. an infinite discontinuity at x=5
possibles answers:
a) I only
b) II only
c) III only
d) I, II, III
Thanks in advance
The denominator cannot be equal to zero, and it is possible when the x-value is,
5x - 25 = 0
5x = 25
x = 5
This is however a removable discontinuity; there is a common factor in the numerator and denominator and therefore the function can be simplified as,
f(x) = (x^2-25)/(5x-25)
f(x) = (x+5)(x-5) / (5)(x-5)
f(x) = (x+5) / 5
where x = all real numbers except 5.
To determine which options must be true for the graph of the function f(x) = (x^2 - 25)/(5x - 25), we need to analyze the behavior of the function as x approaches 5.
First, let's factor the numerator and denominator to help us understand the function better:
f(x) = [(x - 5)(x + 5)] / [5(x - 5)]
Now, we can observe the behavior of the function.
I. A removable discontinuity occurs when there is a hole in the graph at a particular point. This happens when there is a factor common to both the numerator and the denominator that can be canceled out. In this case, we can cancel out the factor (x - 5) in both the numerator and the denominator. As a result, we will have a hole in the graph at x = 5. So, I is correct.
II. A vertical asymptote occurs when the denominator becomes zero and the numerator does not. Looking at the denominator, if we set it equal to zero, we get:
5x - 25 = 0
5x = 25
x = 5
Notice that in this case, the numerator is also zero at x = 5. Therefore, there is no vertical asymptote at x = 5. So, II is incorrect.
III. An infinite discontinuity (also known as a vertical jump) occurs when the function becomes infinite at a certain point. To investigate this, let's observe the limit of the function as x approaches 5 from both sides.
As x approaches 5 from the left (x < 5):
lim [(x^2 - 25)/(5x - 25)] as x approaches 5^- = (-∞)/0 = -∞
As x approaches 5 from the right (x > 5):
lim [(x^2 - 25)/(5x - 25)] as x approaches 5^+ = (∞)/0 = ∞
Since the function approaches different infinite values from each side, there is an infinite discontinuity at x = 5. Therefore, III is correct.
Based on this analysis, the correct answer is d) I, II, III.
To determine whether there is a removable discontinuity, a vertical asymptote, or an infinite discontinuity at x=5 for the function f(x) = (x^2-25)/(5x-25), we need to investigate the behavior of the function near this value.
First, let's consider the behavior as x approaches 5 from both sides. We can simplify the function to make it easier to interpret:
f(x) = (x^2-25)/(5x-25)
= (x+5)(x-5)/(5(x-5))
= (x+5)/5
Now, as x approaches 5, the function becomes f(x) = (5+5)/5 = 2. Therefore, there is no hole or removable discontinuity at x=5 because the function is defined and continuous at that point.
Next, let's consider the limit of the function as x approaches 5 from both sides. Taking the limit, we get:
lim(x->5-) (x+5)/5 = (5+5)/5 = 2
lim(x->5+) (x+5)/5 = (5+5)/5 = 2
Since the limit is the same from both sides and it equals the value of the function at x=5, there is no vertical asymptote at x=5.
Furthermore, the function is defined and continuous at x=5, so there is no infinite discontinuity.
Therefore, the correct answer is:
a) I only