Determining Points of Discontinuity

[swrivas]

Determine all points of discontinuity for f(x)= x²-2, x>1; 2x-3, x<=1

I tried making sense of it by just drawing the two functions of f(x), but I'm not sure how to find the points of discontinuity for these. From what it looks like, there's a parabola with a tangent line running along it. Any ideas on how to prove/determine the points of discontinuity for this mathematically?

Another one that tripped me up:

Let f(x)= sqrtx²-4/sqrtx-2. Find the value that should be assigned to f(2) to guarantee that f will be continuous at x=2.

Should I just start by simplifying the function? I end up with sqrt(x+2) but I'm not sure where to go from there or if that's even a step in the right direction.

 

[Deleted member 4993]

Determine all points of discontinuity for f(x)= x²-2, x>1; 2x-3, x<=1

I tried making sense of it by just drawing the two functions of f(x), but I'm not sure how to find the points of discontinuity for these. From what it looks like, there's a parabola with a tangent line running along it. Any ideas on how to prove/determine the points of discontinuity for this mathematically?

Another one that tripped me up:

Let f(x)= sqrt(x²-4)/sqrt(x-2). Find the value that should be assigned to f(2) to guarantee that f will be continuous at x=2.

Should I just start by simplifying the function? I end up with sqrt(x+2) but I'm not sure where to go from there or if that's even a step in the right direction.

What is the definition of discontinuity of a function?

For the second problem - again definition of continuity comes in - f(2) is just √(x+2) = √(2+2) = 2

And do not forget the grouping symbols () - without those the meaning of the problems change.

 

[swrivas]

What is the definition of discontinuity of a function?

For the second problem - again definition of continuity comes in - f(2) is just √(x+2) = √(2+2) = 2

And do not forget the grouping symbols () - without those the meaning of the problems change.

From my understanding, it's discontinuous anywhere there is a break in the function. Either by removable discontinuity, jumps, asymptotes, etc. So would it just be discontinuous where x=1?

 

[JeffM]

From my understanding, it's discontinuous anywhere there is a break in the function. Either by removable discontinuity, jumps, asymptotes, etc. So would it just be discontinuous where x=1?

The definition is

\(\displaystyle f(x)\ is\ everywhere\ continuous\ if\ and\ only\ if\ (1)\ f(a) \in \mathbb R\ for\ any\ a \in \mathbb R,\ and\ (2)\ \displaystyle \lim_{x \rightarrow a^+}f(x) = f(a) = \lim_{x \rightarrow a^-}f(x)\ for\ any\ a \in \mathbb R.\)

So, for your first problem, ask these questions:

If x < 1, is f(x) continuous? In a very formal course, you might have to demonstrate the answer to this question, but it is so obvious that in most courses you would not have to justify your answer.

If x > 1, is f(x) continuous? Same point as above.

As you saw yourself, the more serious questions arise when x = 1.

Is f(1) a real number?

What is f(1)?

What is the left limit as f(x) approaches 1?

What is the right limit as f(x) approaches 1?

So is f(x) continuous at x = 1?

So is f(x) everywhere continuous or not?

 

[swrivas]

The definition is

\(\displaystyle f(x)\ is\ everywhere\ continuous\ if\ and\ only\ if\ (1)\ f(a) \in \mathbb R\ for\ any\ a \in \mathbb R,\ and\ (2)\ \displaystyle \lim_{x \rightarrow a^+}f(x) = f(a) = \lim_{x \rightarrow a^-}f(x)\ for\ any\ a \in \mathbb R.\)

So, for your first problem, ask these questions:

If x < 1, is f(x) continuous? In a very formal course, you might have to demonstrate the answer to this question, but it is so obvious that in most courses you would not have to justify your answer.

If x > 1, is f(x) continuous? Same point as above.

As you saw yourself, the more serious questions arise when x = 1.

Is f(1) a real number?

What is f(1)?

What is the left limit as f(x) approaches 1?

What is the right limit as f(x) approaches 1?

So is f(x) continuous at x = 1?

So is f(x) everywhere continuous or not?

The concepts are still fairly new so I apologize for being dense. The limits are obvious and easy if I am already looking at a graph of the function(s), but what's giving me trouble is having to see it (without the graphs being provided) just by looking at the functions. If I plug in 1 for x, both functions give me f(x)= -1. What should I take away from this?

 

[Deleted member 4993]

The concepts are still fairly new so I apologize for being dense. The limits are obvious and easy if I am already looking at a graph of the function(s), but what's giving me trouble is having to see it (without the graphs being provided) just by looking at the functions. If I plug in 1 for x, both functions give me f(x)= -1. What should I take away from this?

That f(x) reaches the same value from the left and from the right → f(1^-) = -1 = f(1⁺) = f(1)

So the function is continuous at x = 1 (although there is a "knee" at x =1)

 

[JeffM]

The concepts are still fairly new so I apologize for being dense. The limits are obvious and easy if I am already looking at a graph of the function(s), but what's giving me trouble is having to see it (without the graphs being provided) just by looking at the functions. If I plug in 1 for x, both functions give me f(x)= -1. What should I take away from this?

As Subhotosh Khan has already explained, the limits from left and right are equal to the value of the function at x = 1 so the function is continuous there. That in turn makes it continuous everywhere. BUT, and this is why this kind of problem is given in calculus, the slopes are not the same on either side of 1. That is what SK meant when he said there was a "knee" there. As you will soon find, that means that this function is not differentiable at x = 1.

I am skipping ahead, but if f(x) is differentiable at a, it is also continuous. However, if g(x) is continuous at at b, it may or may not be differentiable at b.

Differentiable functions, which are what you study in differential calculus, are not just continuous, but their slopes can also be described a a continuous function. They are very well behaved functions that change "smoothly." They are sort of super-continuous.