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QuestionFind values of xx where f(x)={x2−1,x<1x,x≥1f(x)=\begin{cases}x^{2}-1, & x<1 \\ x, & x \geq 1\end{cases} is discontinuous and explain why.

Studdy Solution

STEP 1

Assumptions1. We have a piecewise function f(x)f(x) defined as f(x)=x−1f(x) = x^ -1 for x<1x <1 and f(x)=xf(x) = x for x≥1x \geq1. . We need to find all values of xx where f(x)f(x) is discontinuous.
3. A function is discontinuous at a point if the left-hand limit, right-hand limit, or the function value at that point do not exist or are not equal.

STEP 2

We first consider the point x=1x =1. We need to find the left-hand limit, right-hand limit, and the function value at x=1x =1.
The left-hand limit is given by lim⁡x→1−f(x)\lim_{x \to1^-} f(x), the right-hand limit is given by lim⁡x→1+f(x)\lim_{x \to1^+} f(x), and the function value at x=1x =1 is f(1)f(1).

STEP 3

Let's first compute the left-hand limit at x=1x =1.
lim⁡x→1−f(x)=lim⁡x→1−(x2−1)\lim_{x \to1^-} f(x) = \lim_{x \to1^-} (x^2 -1)

STEP 4

Calculate the left-hand limit.
lim⁡x→1−f(x)=(12−1)=0\lim_{x \to1^-} f(x) = (1^2 -1) =0

STEP 5

Now, let's compute the right-hand limit at x=1x =1.
lim⁡x→1+f(x)=lim⁡x→1+x\lim_{x \to1^+} f(x) = \lim_{x \to1^+} x

STEP 6

Calculate the right-hand limit.
lim⁡x→1+f(x)=1\lim_{x \to1^+} f(x) =1

STEP 7

Now, let's compute the function value at x=1x =1.
f(1)=1f(1) =1

STEP 8

Compare the left-hand limit, right-hand limit, and function value at x=1x =1.
The left-hand limit is 00, the right-hand limit and function value are both 11.
Since the left-hand limit does not equal the right-hand limit, the function is discontinuous at x=1x =1.

STEP 9

The reason for the discontinuity at x=x = is that the left-hand limit and right-hand limit are not equal. This is a type of discontinuity known as a jump discontinuity.
The function f(x)f(x) is discontinuous at x=x = due to a jump discontinuity.

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