1. ## Anwer check for continuous question please

If f(x)= { abs(x-3)/(x-3) if x does not =3 then f(x) is:
1 if x=3

a) Undefined at x=3
B) continuous from left at x=3
c) Continuous at x=3 ( from both sides)
d) continuous fro right at x=3
e) differentiable at x=3

The answer would be a) correct? since it is disscontinuous at 3 for abs(x-3)/(x-3)?

2. Is this the function?
$\displaystyle f(x)=\begin{cases} \dfrac{|x - 3|}{x - 3} & \text{if}\:x \ne 3\\\\ 1 & \text{if}\:x = 3 \end{cases}$

If so, then no, the answer is not (a). f(x) is defined at x = 3. The second part of the piecewise definition tells you this: "1 if x = 3", so f(3) = 1.
The answer is (d). The function is continuous for all $\displaystyle x \ge 3$.

It's not (b) because for all x < 3, f(x) = -1, but at x = 3, the value of f(x) jumps to 1.

Since f(x) is continuous from the right but not from the left, the answer cannot be (c).

Differentiability implies continuity, so it can't be (e).

3. Originally Posted by eumyang
Is this the function?
$\displaystyle f(x)=\begin{cases} \dfrac{|x - 3|}{x - 3} & \text{if}\:x \ne 3\\\\ 1 & \text{if}\:x = 3 \end{cases}$

If so, then no, the answer is not (a). f(x) is defined at x = 3. The second part of the piecewise definition tells you this: "1 if x = 3", so f(3) = 1.
The answer is (d). The function is continuous for all $\displaystyle x \ge 3$.

It's not (b) because for all x < 3, f(x) = -1, but at x = 3, the value of f(x) jumps to 1.

Since f(x) is continuous from the right but not from the left, the answer cannot be (c).

Differentiability implies continuity, so it can't be (e).
Ohh ok but how do we know if it comes from the right or left without looking at the sketch?
Like i thought it would be disscontinuous at x=3 since x can not = 3 in the denominator. , so im assuming thats y its coming only from the right. But how did u know its the right not left persay?

4. Originally Posted by kensington
Ohh ok but how do we know if it comes from the right or left without looking at the sketch?
Like i thought it would be disscontinuous at x=3 since x can not = 3 in the denominator. , so im assuming thats y its coming only from the right. But how did u know its the right not left persay?
"Undefined" and "discontinuous" do not mean the same thing. It is possible for a function to not be continuous for some x value(s) but still be defined. The function you gave us is a prime example of that.

If you looked at this function by itself
$\displaystyle f(x) = \dfrac{|x - 3|}{x - 3}$
then yes, it is undefined at x = 3 AND there is a discontinuity at x = 3. But that's not the function you gave, was it? Yours was defined piecewise:
$\displaystyle f(x)=\begin{cases} \dfrac{|x - 3|}{x - 3} & \text{if}\:x \ne 3\\\\ 1 & \text{if}\:x = 3 \end{cases}$

This function is DEFINED at x = 3 (f(3) = 1), but it is still discontinuous at x = 3.

To find out if the graph was continuous from one side or the other, I plugged in values around 3 for x in my head. If you let x = 4, 5, 6, etc. then f(x) = 1. If you let x = 2, 1, 0, etc. then f(x) = -1. Since f(3) also equals 1, I concluded that the graph is continuous from the right of x = 3.

I can make a simple little change to your function and make the graph continuous from the LEFT of x = 3. Here it is; notice the difference?
$\displaystyle f(x)=\begin{cases} \dfrac{|x - 3|}{x - 3} & \text{if}\:x \ne 3\\\\ -1 & \text{if}\:x = 3 \end{cases}$
I redefined f(x) at x = 3 to be -1 instead of 1. Since all of the f(x) values to the left of x = 3 were also -1, we can say here that the graph is continuous from the left of x = 3.