Question

Which one or ones of the following statements is/are true?

If the line x = 1 is a vertical asymptote of y = f(x), then f is not defined at x = 1.
If f(2) > 0 and f(4) < 0, then there exists a number c between 2 and 4 such that f(c) = 0.
If f is continuous at x = 5 and f(5) = 2 and f(4) = 3, then the limit as x approaches 2 of f of the quantity 4 times x squared minus 11 equals 2.
I only

II only

III only

All statements are true.

Answer 1

I know that second one isn't true bc the problem doesn't state that f(x) is necessarily continuous

Answer 2

The last one is \[\Large \lim_{x \rightarrow 2}f(4x^2-11)=2\]

Answer 3

@tkhunny any help please?

Answer 4

I know I is true because that's the definition of a vertical asymptote, it gets close to this number as it approaches 1 from both sides, but doesn't get to it (so it isn't defined)

Answer 5

Asymptote is good.

On the third, have you considered the limit of 4x^2 - 11 as x approaches 2?

Answer 6

I just don't know about the last one, because it never told me about the behavior of the graph at x=2

Answer 7

Well any polynomial function is continuous so it would be defined at 4(2)^2-11 and that would give you f(5) which is 2 so you can conclude that's true?

Answer 8

Wait no, it doesn't give me the option of (I, III) but I know that II is defintely wrong and I is defintely right so it has to be just I, but with that being said, how come III is wrong

Answer 9

Yes. As a general rule (an important programming rule), is that anywhere you can put a value, you should be able to put a function that can be evaluated to that value. This is all we are doing.

Is it the same \(2^{+}\;and\;2^{-}\)?

Answer 10

Yes, they both tend to 5

Answer 11

And it's continuous. I'm not seign the problem. Either the problem is wrong or someone will have to explain it to me.