Hey there! Sign in to join this conversationNew here? Join for free

Is the range correct for this piecewise function? Watch

    • Thread Starter
    Offline

    14
    ReputationRep:
    this latex better work
    this meaty bracket won't come out .__.

    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 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
    Offline

    11
    ReputationRep:
    (Original post by will'o'wisp2)
    i get as the inverse 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.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by atsruser)
    That looks correct to me.
    can you help with the latex formatting stuff?
    Offline

    19
    ReputationRep:
    (Original post by will'o'wisp2)
    x

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



    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?
    Offline

    11
    ReputationRep:
    (Original post by will'o'wisp2)
    this latex better work
    this meaty bracket won't come out .__.

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

    My latex was:

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

    [latex]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}[/latex]

    Note also that you can use [tex]...[/tex] to save typing.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by Desmos)
    f(x)= \begin{pmatrix}{ \dfrac{5}{2}-x\ \text{if}\ x&lt;2} \\ {\dfrac{2}{x^2}\ \text{if}\ x\geq 2 }\end



    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:

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

    [latex]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}[/latex]

    Note also that you can use [tex]...[/tex] to save typing.
    coolio thanks both
    Offline

    10
    ReputationRep:
    Call me wierd but I just want to be sure.

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

    14
    ReputationRep:
    (Original post by simon0)
    Call me wierd but I just want to be sure.

    Should the range of the inverse ( f^{-1} ) be the same as the domain of the orginal function ( f ) 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
    Offline

    10
    ReputationRep:
    (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} ) be:

    

f^{-1}(x) = \begin{cases}

\frac{5}{2} - x, & \text{with range}, \, (-\infty, 2),\\

\sqrt{ \frac{2}{x} }, & \text{with range}, \, [2, \infty).

\end{cases}
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by simon0)
    Sorry, sticking my head out, should the range of the inverse function ( f^{-1} ) be:

     

f^{-1}(x) = 

\begin{cases} 

\frac{5}{2} - x, & \text{with range}, \, (-\infty, 2)\\ 

\sqrt{ \frac{2}{x} }, & \text{with range}, \, (2, \infty). 

\end{cases} 

.
    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
    Offline

    10
    ReputationRep:
    (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} ).

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

    

f(x) \, = \, \begin{cases} \frac{5}{2} - x, & \text{where} \, x &lt; 2, \, \text{has range} \, (\frac{1}{2}, \infty),\\

\frac{2}{x^{2}}, & \text{where} \, x \geq 2, \, \text{has range} \, (0, \frac{1}{2}]. \end{cases}

    I promise this is my final objection and I apologise for the questions, I just want to make sure I understood your question correctly.
    • Thread Starter
    Offline

    14
    ReputationRep:
    (Original post by simon0)
    Okay ,sorry, I misunderstood the question. :-)

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

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

    

f(x) \, = \, \begin{cases} \frac{5}{2} - x, & \text{where} \, x &lt; 2, \, \text{has range} \, (\frac{1}{2}, \infty),\\

\frac{2}{x^{2}}, & \text{where} \, x \geq 2, \, \text{has range} \, (0, \frac{1}{2}]. \end{cases}

    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
    Offline

    10
    ReputationRep:
    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
     
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    (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.

    \frac{5}{2}-x for x &lt; 2 has range (\frac{1}{2}, \infty)
    \frac{2}{x^2} for x \geq 2 has range (0,\frac{1}{2}]

    Then the overall range of the function would be (0,\frac{1}{2}] \cup (\frac{1}{2}, \infty) = (0, \infty)
    • Thread Starter
    Offline

    14
    ReputationRep:
    (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.

    \frac{5}{2}-x for x &lt; 2 has range (\frac{1}{2}, \infty)
    \frac{2}{x^2} for x \geq 2 has range (0,\frac{1}{2}]

    Then the overall range of the function would be (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 -__________-
    Offline

    11
    ReputationRep:
    (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
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.