Question

Find all values of `x` where the following function is discontinuous? `f(x) = {(1-x, "if " x<1), (2, "if " 1 <= x <= 2), (4-x, "if " x>2):}`

Is the reasoning below right? I got this question from Alamo's Website but the answer was contradicting (they claim 1,2 are answers, prove otherwise and do not provide for negative numbers).

This expression is

continuous at $x < 1$

continuous at $x > 2$

continuous at $x = 2$

It is discontinuous at $x=1$

1. Test $x \to 1$

Test Rule #1 test if $f(1)$ exists

When x is between 1 and 2 we will use the function $f(x) = 2$.

or $f(1) = 2$

therefore the function is defined at x = 1 and passes the first test of continuity.

Test Rule #2, test if $\lim\limits_{x \to 1} f(x)$ exists

$\lim\limits_{x \to 1^+} f(x) = 2$

$\lim\limits_{x \to 1^-} f(x) = 1-x = 0$

Compare the one-sided limits

$0 \ne 2$

Since the test fails in Test Rule `2**, we do not check **Test Rule`3 ... $f(x) = f(c)$

$\therefore$ DISCONTINUOUS at x = 1

2. Test $x \to 2$

Test Rule #1 test if $f(c)$ or $f(2)$ exists

$f(2) = 2$ therefore the function is defined at $x = 2$ and passes the First test of continuity.

Test Rule #2 Test if $\lim\limits_{x \to 2} f(x)$ exists

$\lim\limits_{x \to 2^-} f(x) = 2$

$\lim\limits_{x \to 2^+} f(x) = (4-x) = (4-2) = 2$

By comparing the one-sided limits $2 = 2$, we find that this passes the Second test of continuity.

Since $\lim\limits_{x \to 2} f(x) = f(c)$, this expression passes the Third test of continuity.

$\therefore$ CONTINUOUS at $x = 2$

3. Test $x \to n$ where $n > 2$

For all $n > 2$

Test Rule #1 test if $f(c)$ or $f(n)$ exists

$f(n) = (4-n)$ therefore the function is defined at $x = n$ and passes the First test of continuity.

Test Rule #2 Test if $\lim\limits_{x \to n} f(x)$ exists

$\lim\limits_{x \to n^-} f(x) = (4-x) = (4-n)$
$\lim\limits_{x \to n^+} f(x) = (4-x) = (4-n)$

By comparing the one-sided limits, we find that this passes the Second test of continuity.

Since $\lim\limits_{x \to n} f(x) = f(c)$, this expression passes the Third test of continuity.

$\therefore$ CONTINUOUS at $x > 2$

4. Test $x \to n$ where $n < 1$

For all $n < 1$

Test Rule #1 test if $f(c)$ or $f(n)$ exists

$f(x) = (1-x)$ therefore the function is defined at $x = n$ and passes the First test of continuity.

Test Rule #2 Test if $\lim\limits_{x \to n} f(x)$ exists

$\lim\limits_{x \to n^-} f(x) = (1-x) = (1-n)$
$\lim\limits_{x \to n^+} f(x) = (1-x) = (1-n)$

By comparing the one-sided limits, we find that this passes the Second test of continuity.

Since $\lim\limits_{x \to n} f(x) = f(c)$, this expression passes the Third test of continuity.

$\therefore$ CONTINUOUS at $x < 1$

Answer 1

`x=1` is point of discontinuity

Explanation:

There is single point of discontinuity `x=1` where given function `f(x)` is discontinuous because

`LHL=\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{-}}(1-x)=\lim_{h\to 0}(1-(1-h))=\lim_{h\to 0}h=0`

`RHL=\lim_{x\to0^+}f(x)=2`

`f(2)=2`

`\therefore LHL\neRHL=f(2)=2`