Question

Pls help this is pretty urgent

Answer 1

See explanation

Explanation:

a. f(2)=ln|2-1|/2

f(2)=ln(1)/2

f(2)=0/2

f(2)=0

b. x can't equal 0 or 1. ln(x-1) has no solutions when x=1 and when the denominator is 0, there is no solution.

c. Nonremovable. There is no way to simply f(x) any further. If there was, there would be a point of discontinuity and it would be removable.

Answer 2

(a) See below.

(b) x = 0 or x = 1

(c) x = 0 removable, x = 1 non-removable

Explanation:

Given rational function:

`f(x)=(\ln |x-1|)/(x)`

Part (a)

Substitute x = 2 into the given rational function:

`\begin{aligned}\implies f(2) & =(\ln |2-1|)/(2)\\\\ & =(\ln 1)/(2)\\\\ & =(0)/(2)\\\\ & =0\end{aligned}`

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

Part (b)

Given interval: [-2, 2]

Logs of negative numbers or zero are undefined. As the numerator is the natural log of an absolute value, 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].

Part (c)

x = 1 is a vertical asymptote. As the function exists on both sides of this vertical asymptote, it is an infinite discontinuity. Since the function doesn't approach a particular finite value, the limit does not exist. Therefore, x = 1 is a non-removable discontinuity.

A hole 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.

`\implies f(0)=(\ln |0-1|)/(0)=(\ln 1)/(0)=(0)/(0)`

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.