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Is the range correct for this piecewise function?

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

Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ if\ x<2} \\ {\dfrac{2}{x^2}\ if\ x\geq 2 }\end




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


i get as the inverse
Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ (-\infty ,\dfrac{1}{2})} \\ {\sqrt \dfrac{2}{x} [\dfrac{1}{2}, \infty) }\end




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
(edited 6 years ago)
Original post by will'o'wisp2

i get as the inverse
Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ (-\infty ,\dfrac{1}{2})} \\ {\sqrt \dfrac{2}{x} [\dfrac{1}{2}, \infty) }\end



That looks correct to me.
Original post by atsruser
That looks correct to me.


can you help with the latex formatting stuff?
Reply 3
Avatar for MR1999
MR1999
21
Original post by will'o'wisp2
x



Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ \text{if}\ x<2} \\ {\dfrac{2}{x^2}\ \text{if}\ x\geq 2 }\end





Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ (-\infty ,\dfrac{1}{2}\right{)}} \\ {\sqrt{\dfrac{2}{x}} \text{ } \left[\dfrac{1}{2}, \infty) }\end







Does it look good now?
(edited 6 years ago)
Original post by will'o'wisp2
this latex better work
this meaty bracket won't come out .__.

Unparseable latex formula:

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



Unparseable latex formula:

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}





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

My latex was:
[noparse]
Unparseable latex formula:

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



Unparseable latex formula:

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}


[/noparse]
Note also that you can use [noparse]......[/noparse] to save typing.
Original post by Desmos
Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ \text{if}\ x<2} \\ {\dfrac{2}{x^2}\ \text{if}\ x\geq 2 }\end





Unparseable latex formula:

f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ (-\infty ,\dfrac{1}{2}\right{)}} \\ {\sqrt{\dfrac{2}{x}} \text{ } \left[\dfrac{1}{2}, \infty) }\end







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:
[noparse]
Unparseable latex formula:

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



Unparseable latex formula:

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}


[/noparse]
Note also that you can use [noparse]......[/noparse] to save typing.


coolio thanks both
Reply 6
Avatar for simon0
simon0
13
Call me wierd but I just want to be sure.

Should the range of the inverse (f−1 f^{-1} ) be the same as the domain of the orginal function (f f ) in this case?
(edited 6 years ago)
Original post by simon0
Call me wierd but I just want to be sure.

Should the range of the inverse (f−1 f^{-1} ) be the same as the domain of the orginal function (f f ) in this case?

The graph of the function is below:

Attachment not found

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
Reply 8
Avatar for simon0
simon0
13
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 (f−1 f^{-1} ) be:

Unparseable latex formula:

[br]f^{-1}(x) = \begin{cases}[br]\frac{5}{2} - x, & \text{with range}, \, (-\infty, 2),\\[br]\sqrt{ \frac{2}{x} }, & \text{with range}, \, [2, \infty).[br]\end{cases}[br]

(edited 6 years ago)
Original post by simon0
Sorry, sticking my head out, should the range of the inverse function (f−1 f^{-1} ) be:

[br]f−1(x)=[br]{[br]52−x,with range, (−∞,2)[br]2x,with range, (2,∞).[br][br] [br]f^{-1}(x) = [br]\begin{cases} [br]\frac{5}{2} - x, & \text{with range}, \, (-\infty, 2)\\ [br]\sqrt{ \frac{2}{x} }, & \text{with range}, \, (2, \infty). [br]\end{cases} [br].


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
Reply 10
Avatar for simon0
simon0
13
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 f−1 f^{-1} ).

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

Unparseable latex formula:

[br]f(x) \, = \, \begin{cases} \frac{5}{2} - x, & \text{where} \, x < 2, \, \text{has range} \, (\frac{1}{2}, \infty),\\[br]\frac{2}{x^{2}}, & \text{where} \, x \geq 2, \, \text{has range} \, (0, \frac{1}{2}]. \end{cases}[br]



I promise this is my final objection and I apologise for the questions, I just want to make sure I understood your question correctly.
(edited 6 years ago)
Original post by simon0
Okay ,sorry, I misunderstood the question. :-)

(I thought you were referring to the range of the inverse function f−1 f^{-1} ).

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

Unparseable latex formula:

[br]f(x) \, = \, \begin{cases} \frac{5}{2} - x, & \text{where} \, x < 2, \, \text{has range} \, (\frac{1}{2}, \infty),\\[br]\frac{2}{x^{2}}, & \text{where} \, x \geq 2, \, \text{has range} \, (0, \frac{1}{2}]. \end{cases}[br]



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
.


[QUOTE="Desmos;74085784"]??

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
Reply 12
Avatar for simon0
simon0
13
I created a graph of the piecewise function beforehand just to make sure but we can see the range of the piecewise function:
Graph.png49.1KB
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.

52−x\frac{5}{2}-x for x<2x < 2 has range (12,∞)(\frac{1}{2}, \infty)
2x2\frac{2}{x^2} for x≥2x \geq 2 has range (0,12](0,\frac{1}{2}]

Then the overall range of the function would be (0,12]∪(12,∞)=(0,∞)(0,\frac{1}{2}] \cup (\frac{1}{2}, \infty) = (0, \infty)
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.

52−x\frac{5}{2}-x for x<2x < 2 has range (12,∞)(\frac{1}{2}, \infty)
2x2\frac{2}{x^2} for x≥2x \geq 2 has range (0,12](0,\frac{1}{2}]

Then the overall range of the function would be (0,12]∪(12,∞)=(0,∞)(0,\frac{1}{2}] \cup (\frac{1}{2}, \infty) = (0, \infty)


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 -__________-
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 :smile:

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