Calc Problem with Functions

[flin5845]

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For which values of x is the function f(x) undefined if f(x)=(x+3)/(x^2-9)?

I feel super stupid because I do know really remember what it means if a function is undefined.

I am assuming is that it is f(x)=0 but I am not sure if that is correct.

Can someone explain to me what it means for a function to be undefined?

 

[AlbinoPolarBear]

For which values of x is the function f(x) undefined if f(x)=(x+3)/(x^2-9)?

I feel super stupid because I do know really remember what it means if a function is undefined.

I am assuming is that it is f(x)=0 but I am not sure if that is correct.

Can someone explain to me what it means for a function to be undefined?

Any value divided by 0 is undefined.

So you just expand the bottom, simplify, and solve for denominator = 0.

f(x) = (x+3)/(x^2-9)
f(x) = (x+3)/ [(x+3)(x-3)]
f(x) = 1/(x-3)

If you set x=3, then f(x) = 1/0 = undefined.

 

[flin5845]

Any value divided by 0 is undefined.

So you just expand the bottom, simplify, and solve for denominator = 0.

f(x) = (x+3)/(x^2-9)
f(x) = (x+3)/ [(x+3)(x-3)]
f(x) = 1/(x-3)

If you set x=3, then f(x) = 1/0 = undefined.

Close....but simplifying this will give you the wrong answer, Since X=-3...when you simplify you are changing the value function.

This problem is from the Math Destroyer Practice test 4 #28

 

[AlbinoPolarBear]

Close....but simplifying this will give you the wrong answer, Since X=-3...when you simplify you are changing the value function.

This problem is from the Math Destroyer Practice test 4 #28

I came back to this thread to see if you thanked me yet.. butttt... 😛

I'm not close, I'm right on 😉
Simplifying will give you the right answer.
That's why you simplify... to get rid of common multiples on the top and bottom.
The function will be undefined at x=3...

Don't have math destroyer.

 

[rockclock]

I came back to this thread to see if you thanked me yet.. butttt... 😛

I'm not close, I'm right on 😉
Simplifying will give you the right answer.
That's why you simplify... to get rid of common multiples on the top and bottom.
The function will be undefined at x=3...

Don't have math destroyer.

The function is undefined at x = 3 because there's a vertical asymptote. The lim of f(x) as x tends to 3 also does not exist because it's different when approaching from either side.

The function is undefined at x = -3 as a single discontinuity. The lim of f(x) as x tends to -3 exists (it equals -1/6) by Albino's reasoning, but I still say the original function is undefined there. It and Albino's simplified version are identical functions EXCEPT for at x = -3.

 

[AlbinoPolarBear]

The function is undefined at x = 3 because there's a vertical asymptote. The lim of f(x) as x tends to 3 also does not exist because it's different when approaching from either side.

The function is undefined at x = -3 as a single discontinuity. The lim of f(x) as x tends to -3 exists (it equals -1/6) by Albino's reasoning, but I still say the original function is undefined there. It and Albino's simplified version are identical functions EXCEPT for at x = -3.

When they asked for undefined, I solved to get the vertical asymptote, because they are generally one and the same.

I guess I relearned something today...
Vertical asymptote means that the function is undefined, while function undefined does not mean there has to be a vertical asymptote.

If you plug it into you calc, it pretty much shows a continuous line that goes through x = -3, I guess the single discontinuity wouldn't show up as well.

 

[rockclock]

When they asked for undefined, I solved to get the vertical asymptote, because they are generally one and the same.

I guess I relearned something today...
Vertical asymptote means that the function is undefined, while function undefined does not mean there has to be a vertical asymptote.

If you plug it into you calc, it pretty much shows a continuous line that goes through x = -3, I guess the single discontinuity wouldn't show up as well.

Yeah, like I said, I agree that the limit as x tends to that point exists, but I'm pretty sure it would be correct to say that f(x) is undefined there.

http://en.wikipedia.org/wiki/Removable_singularity