Can someone please help me
1 answer:
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Answer:
(a) See below.
(b) x = 0 or x = 1
(c) x = 0 removable, x = 1 non-removable
Step-by-step explanation:
Given rational function:
<u>Part (a)</u>
Substitute x = 2 into the given rational function:
Therefore, as the function is defined at x = 2, the function is continuous at x = 2.
<u>Part (b)</u>
Given interval: [-2, 2]
Logs of negative numbers or zero are undefined. As the numerator is the natural log of an <u>absolute value</u>, the numerator is undefined when:
|x - 1| = 0 ⇒ x = 1.
A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.
So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].
<u>Part (c)</u>
x = 1 is a <u>vertical asymptote</u>. As the function exists on both sides of this vertical asymptote, it is an <u>infinite discontinuity</u>. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.
A <u>hole</u> exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.
Therefore, there is a hole at x = 0.
The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it. Therefore, x = 0 is a removable discontinuity.
