Continuity of Piecewise-Defined Functions

[nycmathdad]

Determine whether f(x) is continuous at x = 3.

I don't know how to write a piecewise function using LaTex.

I will divide the function as shown below.

Top Part

f(x) = (x^2 - 9)/(x - 3) if x < 3

Middle Part

f(x) = 9 if x = 3

Lower Part

f(x) = x^2 - 3 if x > 3

Let me see.

Is f(3) defined?

The middle part tells me that f(3) is 9. So, condition 1 is satisfied.

I now need to take the limit of f(x) as x tends to 3 from the left and right as part of condition2.

From the left, I must use the top part.

f(x) = (x^2 - 9)/(x - 3)

f(x) = (x - 3)(x + 3)/(x - 3)

f(x) = x + 3

Let x = 3.

f(3) = 6

From the right, I evaluate f(x) = x^2 - 3 at 3.

Doing so, I get 9 - 3 or 6.

So, condition 2 is satisfied because LHL = RHL.

But the limit as x tends to 3 is 6 and condition 1 tells me that f(3) = 9. The third condition fails.

So, f(x) is not continuous at x = 3.

Yes? No? Maybe?

 

[Harry_the_cat]

Not continuous is correct.

However just watch the way you write things.

When considering the left limit, you can't say f(3)=6, because f(3)=9,
but you can say \(\displaystyle lim_{x\to 3^-}f(x)=6\).

Same with from the right.

 

[jonah2.0]

Beer soaked ramblings follow.

Determine whether f(x) is continuous at x = 3.

I don't know how to write a piecewise function using LaTex.

I will divide the function as shown below.

Top Part

f(x) = (x^2 - 9)/(x - 3) if x < 3

Middle Part

f(x) = 9 if x = 3

Lower Part

f(x) = x^2 - 3 if x > 3

Let me see.

Is f(3) defined?

The middle part tells me that f(3) is 9. So, condition 1 is satisfied.

I now need to take the limit of f(x) as x tends to 3 from the left and right as part of condition2.

From the left, I must use the top part.

f(x) = (x^2 - 9)/(x - 3)

f(x) = (x - 3)(x + 3)/(x - 3)

f(x) = x + 3

Let x = 3.

f(3) = 6

From the right, I evaluate f(x) = x^2 - 3 at 3.

Doing so, I get 9 - 3 or 6.

So, condition 2 is satisfied because LHL = RHL.

But the limit as x tends to 3 is 6 and condition 1 tells me that f(3) = 9. The third condition fails.

So, f(x) is not continuous at x = 3.

Yes? No? Maybe?

Another repost.

Piecewise-defined Function

Determine whether f(x) is continuous at x = 3. I don't know how to write a piecewise function using LaTex. I will divide the function as shown below. Top Part f(x) = (x^2 - 9)/(x - 3) if x < 3 Middle Part f(x) = 9 if x = 3 Lower Part f(x) = x^2 - 3 if x > 3 Let me see. Is f(3)...

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[nycmathdad]

Not continuous is correct.

However just watch the way you write things.

When considering the left limit, you can't say f(3)=6, because f(3)=9,
but you can say \(\displaystyle lim_{x\to 3^-}f(x)=6\).

Same with from the right.

Can you provide me with two more examples?