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Math Help - evaluate and simplify

  1. #1
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    evaluate and simplify

    hey guys, need help with a practise question... thank you!!

    if
    <br />
f(x) = \frac{2}{3x+4}<br />

    evaluate and simplify

    <br />
\frac{(2+h)-f(2)}{h}<br />
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by jvignacio View Post
    hey guys, need help with a practise question... thank you!!

    if
    <br />
f(x) = \frac{2}{3x+4}<br />

    evaluate and simplify

    <br />
\frac{(2+h)-f(2)}{h}<br />
    have you wriiten the question correctly?

    first, f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}

    so, <br />
\frac{(2+h)-f(2)}{h} = \frac{(2+h) + \frac{1}{5}}{h} = \frac{10 + 5h + 1}{5h} = \frac{11+5h}{5h}<br />


    but this looks familiar.. i was thinking that the problem was

    \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}

    am i right?
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  3. #3
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    Quote Originally Posted by kalagota View Post
    have you wriiten the question correctly?

    first, f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}

    so, <br />
\frac{(2+h)-f(2)}{h} = \frac{(2+h) + \frac{1}{5}}{h} = \frac{10 + 5h + 1}{5h} = \frac{11+5h}{5h}<br />


    but this looks familiar.. i was thinking that the problem was

    \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}

    am i right?
    yes sorry thats it! your right. thats the problem. how would i solve it like that?
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  4. #4
    MHF Contributor kalagota's Avatar
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    it should be familiar to you that

    f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}

    if not yet, all you have to do is to substitute 2+h on x of f(x) to get f(2+h)

    then solve for that limit..

    f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}

    so, \lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}...

    i hope, you can continue from here..
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  5. #5
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    Quote Originally Posted by kalagota View Post
    it should be familiar to you that

    f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}

    if not yet, all you have to do is to substitute 2+h on x of f(x) to get f(2+h)

    then solve for that limit..

    f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}

    so, \lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}...

    i hope, you can continue from here..
    okay i got upto..

    <br />
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}<br />

    am i right? not sure what to do after
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    okay i got upto..

    <br />
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}<br />

    am i right? not sure what to do after

    anyone?
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by jvignacio View Post
    okay i got upto..

    <br />
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}<br />

    am i right? not sure what to do after
    that should be negative..


    <br />
\lim_{h\rightarrow 0} \frac{\frac{-3h}{50+15h}}{h} = \lim_{h\rightarrow 0} \frac{-3h}{50+15h} \cdot \frac{1}{h} =  \lim_{h\rightarrow 0} \frac{-3}{50 + 15h} ...<br />
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by jvignacio View Post
    okay i got upto..

    <br />
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}<br />

    am i right? not sure what to do after
    <br />
-\ \frac{\frac{3h}{50+15h}}{h}=-\ \frac{3}{50+15h}<br />

    Then the limit of this as h goes to 0 is obtained just by putting h=0, so is: -3/50.

    RonL
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