2nd week of calculus 1 homework problem

• September 9th 2008, 06:03 PM
2nd week of calculus 1 homework problem
really easy, but we just started. I think the answer is false. Its worth a lot of points so I am making sure I am right.

f(x)=(x^(2)-1)/(x-1) is continuous on [-2,2].

True or False?
• September 9th 2008, 06:12 PM
Legendsn3verdie
Quote:

really easy, but we just started. I think the answer is false. Its worth a lot of points so I am making sure I am right.

f(x)=(x^(2)-1)/(x-1) is continuous on [-2,2].

True or False?

i THINK it is continuos at that point

cause:

1. x is defined as -1
2. u gotta take the lim from above and below negitive 2.

lim x->-2+ ((-2)^2-1)/((-2)-1)) = -1

lim x->-2- ((-2)^2-1)/((-2)-1)) = -1

it is contiuos

(SOMEONE CHECK THIS)
• September 9th 2008, 06:14 PM
Aryth
For this, to make it false, you have to find one x that makes it discontinuous. This is easy, because it is rational, all you have to do is set the denominator equal to zero and solve for x, therefore:

$x - 1 = 0$

$x = 1$

And at that point you have a discontinuity. It lies in the interval [-2,2], therefore the function is not continuous on that interval.

P.S. Legends, she's using interval notation, so [-2,2] is not a point, but a set of points.
• September 9th 2008, 06:17 PM
Legendsn3verdie
Quote:

Originally Posted by Aryth
For this, to make it false, you have to find one x that makes it discontinuous. This is easy, because it is rational, all you have to do is set the denominator equal to zero and solve for x, therefore:

$x - 1 = 0$

$x = 1$

And at that point you have a discontinuity. It lies in the interval [-2,2], therefore the function is not continuous on that interval.

P.S. Legends, she's using interval notation, so [-2,2] is not a point, but a set of points.

oopps i read it as, is the limit discotinuios at the point (-2,2) so i tested the -2 value.. didnt realize it was a interval maybe i should read better
• September 9th 2008, 06:17 PM
thanks so much. when i graphed it it didn't look like it would be continuous. i don't know if thats a good way of telling but thanks!
• September 9th 2008, 06:24 PM
Aryth
Graphing is a good way to VERIFY your work, but going by the graph alone is not usually a great idea. Always try to arrive at the answer on your own and use the graph to verify your answer.

For your equation, it's pretty simple to further show the discontinuity, here's your original equation:

$f(x) = \frac{x^2 - 1}{x - 1}$

There are two ways to show the discontinuity:

1. Finding the hole:

All you have to do here is try to factor the numerator, to cancel the all or part of the denominator, usually, what cancels is a hole (Not sure on the formalities):

Notice that $(x^2 - 1)$ factors to $(x+1)(x-1)$, when you plug that into your function you get:

$f(x) = \frac{(x+1)(x-1)}{x-1}$

If you cancel the (x-1)'s, you get:

$f(x) = x+1$ But, when you canceled the (x-1)'s, you showed that there was a hole in the graph at $x=1$ and is therefore not continuous.

2. Divide by zero

For this, the easiest solution is to cause the rational function to divide by zero, if the point lies within the given interval, then it is not continuous over that interval. All you have to do is set the denominator equal to zero and solve for x:

$x - 1 = 0$

$x = 1$

Now, you plug it into the function:

For $x = 1$

$f(1) = \frac{1^2 - 1}{1 - 1}$

$f(1) = \frac{0}{0}$

This doesn't equal zero, anything divided by zero is undefined, therefore the function is not continuous at $x=1$.

And there you go.