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Math Help - solving the expression

  1. #1
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    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?
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  2. #2
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    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.

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

    <br />
\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)}
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  3. #3
    Moo
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    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 ?


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

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


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

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

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


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

    \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} \\ \\<br />
&= \frac{\frac{x+2}{(x+h+2)(x+2)}-\frac{x+h+2}{(x+2)(x+h+2)}}{h} \\<br />
&= \frac{\frac{(x+2)-(x+h+2)}{(x+h+2)(x+2)}}{h} \end{aligned}

    \begin{aligned} \frac{g(x+h)-g(x)}{h} &= \frac{\frac{-h}{(x+h+2)(x+2)}}{h} \\ \\<br />
&= \frac{-{\color{red}h}}{{\color{red}h}(x+h+2)(x+2)} \\<br />
&= \frac{-1}{(x+h+2)(x+2)} \end{aligned}
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