# functions and limits

• Jul 30th 2007, 07:44 AM
Dili
functions and limits
I have trouble with this questions

f:[3,infinity)-->R f(x)=x^2 -6x +10

The answer for this (in mcqs) is
f^-1:[1,infi)-->[3,infi), f^-1(y)= 3+ (y-1)^-.5

I do know how to get the f inverse but i do not know how to get the domains and ranges.
Can you explain what is meant from [1,infi)-->[3,infi) ?

2nd question

f(x)={x x is rational
-x x is irrational

g(x)={sinx x is rational
x^3 x is irrational

Can you comment on the existance of limx->0f(x) and lim->0g(x)

Thank you

PS. sorry about the written format. I don't know how to use maths sybols(is there a link which tells how)
• Jul 30th 2007, 09:16 AM
CaptainBlack
Quote:

Originally Posted by Dili
2nd question

f(x)={x x is rational
-x x is irrational

g(x)={sinx x is rational
x^3 x is irrational

Can you comment on the existance of limx->0f(x) and lim->0g(x)

Thank you

PS. sorry about the written format. I don't know how to use maths sybols(is there a link which tells how)

(A)

-x <= f(x) <= x

hence as x -> 0 f(x) is trapped between two continuous function both of
which go to 0, so the limit exists and is 0.

(B)

for |x|<1

-x <= g(x) <= x,

so again the limit of g(x) as x->0 is 0.

RonL
• Jul 31st 2007, 04:28 AM
Dili
Can anyone post an answer to the first question?
• Jul 31st 2007, 04:48 AM
topsquark
Quote:

Originally Posted by Dili
f:[3,infinity)-->R f(x)=x^2 -6x +10

The answer for this (in mcqs) is
f^-1:[1,infi)-->[3,infi), f^-1(y)= 3+ (y-1)^-.5

I do know how to get the f inverse but i do not know how to get the domains and ranges.
Can you explain what is meant from [1,infi)-->[3,infi) ?

You have a function f with a domain $[3, \infty)$ and a given range which is all real numbers.

I don't know what "mcqs" is supposed to be, but it looks like you are trying to find the inverse function.

I'm going to first change the form of f a bit. This is not strictly necessary, but will simplify finding the inverse function.
$f(x) = x^2 - 6x + 10$

$f(x) = (x^2 - 6x) + 10$

$f(x) = (x^2 - 6x + 9) - 9 + 10$

$f(x) = (x - 3)^2 + 1$

Now to find the inverse function. I'm going to let y = f(x) represent the function. Then x = f(y) represents the inverse function. So I'm going to solve x = f(y) for y:
$x = (y - 3)^2 + 1$

$(y - 3)^2 = x - 1$

$y - 3 = \sqrt{x - 1}$

$y = 3 + \sqrt{x - 1}$

So the function $f^{-1}$ is given by $f^{-1}(y) = 3 + \sqrt{x - 1}$. (I changed the variable back to y to match with your problem. It's a "dummy" variable and can be anything we wish.)

How about domain and range? Well, the domain of f is $[3, \infty)$ so this is the range of $f^{-1}$. The allowed domain of $f^{-1}$ is $[1, \infty)$. This domain gives the exact range as we expect (it must give a range at least as large as the expected range, else we need to modify the domain such that it will.) So the function $f^{-1}$ may be expressed as:
$f^{-1}: [1, \infty) \to [3, \infty) \mapsto f^{-1}(y) = 3 + \sqrt{x - 1}$

-Dan