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# Is the range correct for this piecewise function? watch

1. this latex better work
this meaty bracket won't come out .__.

guess it'll have to do with a ( instead of { can't figure it out

i get as the inverse

jesus christ -___________-

i give up the last bit is supposed to read the square root of 2 over x with the range as being from 1/2 to infinity
2. (Original post by will'o'wisp2)
i get as the inverse
That looks correct to me.
3. (Original post by atsruser)
That looks correct to me.
can you help with the latex formatting stuff?
4. (Original post by will'o'wisp2)
x

Does it look good now?
5. (Original post by will'o'wisp2)
this latex better work
this meaty bracket won't come out .__.

How about that? Pretty good try with the first lot of latex though.

My latex was:

$f(x)= \begin{cases}{ \frac{5}{2}-x & \text{if } x<2} \\ {\frac{2}{x^2} & \text{if } x \geq 2 }\end{cases}$

$f(x)= \begin{cases}{ \frac{5}{2}-x & x \in (-\infty ,\dfrac{1}{2})} \\ { \frac{2}{\sqrt{x}} & x \in [\frac{1}{2}, \infty) }\end{cases}$

Note also that you can use $$...$$ to save typing.
6. (Original post by Desmos)

Does it look good now?
(Original post by atsruser)
How about that? Pretty good try with the first lot of latex though.

My latex was:

$f(x)= \begin{cases}{ \frac{5}{2}-x & \text{if } x<2} \\ {\frac{2}{x^2} & \text{if } x \geq 2 }\end{cases}$

$f(x)= \begin{cases}{ \frac{5}{2}-x & x \in (-\infty ,\dfrac{1}{2})} \\ { \frac{2}{\sqrt{x}} & x \in [\frac{1}{2}, \infty) }\end{cases}$

Note also that you can use $$...$$ to save typing.
coolio thanks both
7. Call me wierd but I just want to be sure.

Should the range of the inverse () be the same as the domain of the orginal function () in this case?
8. (Original post by simon0)
Call me wierd but I just want to be sure.

Should the range of the inverse () be the same as the domain of the orginal function () in this case?

The graph of the function is below:

Attachment 695208
range of the normal function is the same as the domain of the inverse and the domain of the normal function is same as the range of the inverse
9. (Original post by will'o'wisp2)
range of the normal function is the same as the domain of the inverse and the domain of the normal function is same as the range of the inverse
Sorry, sticking my head out, should the range of the inverse function () be:

10. (Original post by simon0)
Sorry, sticking my head out, should the range of the inverse function () be:

.
no? that seems like the wrong range because this is a "split" function so it's different to just a standard function.

In any case the best way i find to work things out is to look at the domain of the original and ask myself what will come out if i put the smallest possible number in the original function and the largest possible number in that function what will happen then? that range is then the domain of the f^-1 stuff
11. (Original post by will'o'wisp2)
no? that seems like the wrong range because this is a "split" function so it's different to just a standard function.

In any case the best way i find to work things out is to look at the domain of the original and ask myself what will come out if i put the smallest possible number in the original function and the largest possible number in that function what will happen then? that range is then the domain of the f^-1 stuff
Okay ,sorry, I misunderstood the question. :-)

(I thought you were referring to the range of the inverse function ).

In this case, for my final question, should the range of be:

I promise this is my final objection and I apologise for the questions, I just want to make sure I understood your question correctly.
12. (Original post by simon0)
Okay ,sorry, I misunderstood the question. :-)

(I thought you were referring to the range of the inverse function ).

In this case, for my final question, should the range of be:

I promise this is my final objection and I apologise for the questions, I just want to make sure I understood your question correctly.
(Original post by atsruser)
.
(Original post by Desmos)
[latex]?
Which is exactly why i asked this question because according to the booklet that i have, simon is correct and i am wrong

the difference is that simon said that the range of the x^2 function was between 0 and 0.5 whereas i said it was between minus infinity and 0.5
13. I created a graph of the piecewise function beforehand just to make sure but we can see the range of the piecewise function:
Attached Images

14. (Original post by will'o'wisp2)
Which is exactly why i asked this question because according to the booklet that i have, simon is correct and i am wrong

the difference is that simon said that the range of the x^2 function was between 0 and 0.5 whereas i said it was between minus infinity and 0.5
Not sure why you mentioned the inverse in your OP.

Simon is correct, and you didn't answer the question by the looks of it unless I missed something.

Just do it by cases.

for has range
for has range

Then the overall range of the function would be
15. (Original post by RDKGames)
Not sure why you mentioned the inverse in your OP.

Simon is correct, and you didn't answer the question by the looks of it unless I missed something.

Just do it by cases.

for has range
for has range

Then the overall range of the function would be
was making some notes and i just wanted to find the domain of the inverse but turns out i got a different answer to the booklet, which simon has gotten but i got - infinity and 1/2 which i now realise s wrong bc the squared returns oonly positive and 0 doesn't work so then 0<x<1/2 gj me i love doing ranges -__________-
16. (Original post by will'o'wisp2)
Which is exactly why i asked this question because according to the booklet that i have, simon is correct and i am wrong

the difference is that simon said that the range of the x^2 function was between 0 and 0.5 whereas i said it was between minus infinity and 0.5
Yep, looks like there's a range problem. Sorry I didn't check that properly last night. All the dodgy latex confused me

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