Results 1 to 3 of 3

Thread: solving the expression

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    69

    solving the expression

    Let function NoFontMessage () {} jsMath = {Parser: {prototype: {environments: {'eqnarray*' :['Array',null,null,'rcl',[5/18,5/18],3,'D']}}}}; g(x)=1/(x+2) Find the expression

    g(x+h)-g(x) / h

    this is what i did:
    g(x+h)-g(x) / h
    =g(x)+g(h)-g(x)/ h
    =1/(x+2)+g(h)-1/(x+2) / h
    =g(h) / h
    = g

    is this right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by lemontea View Post
    Let g(x)=1/(x+2)
    Find the expression


    g(x+h)-g(x) / h

    this is what i did:
    g(x+h)-g(x) / h
    =g(x)+g(h)-g(x)/ h
    =1/(x+2)+g(h)-1/(x+2) / h
    =g(h) / h
    = g

    is this right?
    Unfortunately no.

    $\displaystyle g(x)=\frac1{x+2}~\implies~g(x+h)=\frac1{x+2+h}$

    $\displaystyle
    \frac{\frac1{x+2+h} - \frac1{x+2}}{h} = \frac{\frac{x+2-(x+2+h)}{(x+2)(x+2+h)}}{h} = \frac{\frac{-h}{(x+2)(x+2+h)}}{h} = -\frac1{(x+2)(x+2+h)}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Quote Originally Posted by lemontea View Post
    Let function NoFontMessage () {} jsMath = {Parser: {prototype: {environments: {'eqnarray*' :['Array',null,null,'rcl',[5/18,5/18],3,'D']}}}}; g(x)=1/(x+2) Find the expression

    g(x+h)-g(x) / h

    this is what i did:
    g(x+h)-g(x) / h
    =g(x)+g(h)-g(x)/ h
    =1/(x+2)+g(h)-1/(x+2) / h
    =g(h) / h
    = g

    is this right?
    Erm.. I don't know what to say... In which grade are you ?


    $\displaystyle g(x+h) {\color{red}\neq} g(x)+g(h)$

    And $\displaystyle \frac{g(h)}{h} {\color{red}\neq} g$ (following your reasoning, it'd make g() lol)...


    You want $\displaystyle \frac{g(x+h)-g(x)}{h}$

    $\displaystyle g(x)=\frac{1}{x+2}$

    --> $\displaystyle g({\color{blue}x+h})=\frac{1}{({\color{blue}x+h})+ 2}=\frac{1}{x+h+2}$


    $\displaystyle \frac{g(x+h)-g(x)}{h} = \frac{\frac{1}{x+h+2}-\frac{1}{x+2}}{h} $

    $\displaystyle \begin{aligned} \frac{g(x+h)-g(x)}{h} &= \frac{\frac{1}{x+h+2} \cdot \frac{x+2}{x+2}-\frac{1}{x+2} \cdot \frac{x+h+2}{x+h+2}}{h} \\ \\
    &= \frac{\frac{x+2}{(x+h+2)(x+2)}-\frac{x+h+2}{(x+2)(x+h+2)}}{h} \\
    &= \frac{\frac{(x+2)-(x+h+2)}{(x+h+2)(x+2)}}{h} \end{aligned}$

    $\displaystyle \begin{aligned} \frac{g(x+h)-g(x)}{h} &= \frac{\frac{-h}{(x+h+2)(x+2)}}{h} \\ \\
    &= \frac{-{\color{red}h}}{{\color{red}h}(x+h+2)(x+2)} \\
    &= \frac{-1}{(x+h+2)(x+2)} \end{aligned}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help solving algebraic expression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 5th 2011, 09:33 AM
  2. [SOLVED] Solving Expression
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Aug 30th 2010, 11:29 AM
  3. Solving for variable in expression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 28th 2009, 05:52 PM
  4. Replies: 7
    Last Post: Aug 16th 2008, 06:05 AM
  5. Replies: 3
    Last Post: Jan 26th 2008, 11:16 AM

Search Tags


/mathhelpforum @mathhelpforum