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Thread: Did something wrong in my calculations, but I don't know what.

  1. #1
    Junior Member ImaCowOK's Avatar
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    Did something wrong in my calculations, but I don't know what.

    \frac{\frac{5}{6(x+h)+3}-\frac{5}{6x+3}}{h}

    Should end up in the form \frac{A}{(Bx+Ch+3)(Dx+3)}

    I got my answer in that form and the values are correct for A, B, C, and D. But when I use it to answer the rest of the questions I'm getting wrong answers. I think I did something wrong in the beginning when getting it in that form.

    Can someone go through the steps and give me your answer to check with mine. I need to see what I did wrong.

    A = -30
    B = 6
    C = 6
    D = 6

    Can you display your final answer in the form: \frac{A}{(Bx+Ch+3)(Dx+3)}
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  2. #2
    MHF Contributor Unknown008's Avatar
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    It seems good to me...

    I'll take the numerator first.

    \frac{5}{6(x+h) + 3} - \frac{5}{6x+3} = \frac{5}{6x+6h + 3} - \frac{5}{6x+3}

    = \frac{5(6x+3) - 5(6x+6h+3)}{(6x+6h + 3)(6x+3)}

    = \frac{(30x+15) - (30x+30h+15)}{(6x+6h + 3)(6x+3)}

    = \frac{30x+15 - 30x-30h-15}{(6x+6h + 3)(6x+3)}

    = \frac{-30h}{(6x+6h + 3)(6x+3)}

    Now, the new fraction:

    \frac{\frac{5}{6(x+h) + 3} - \frac{5}{6x+3}}{h} = \frac{\frac{-30h}{(6x+6h + 3)(6x+3)}}{h}

    = \frac{-30h}{h(6x+6h + 3)(6x+3)}

    = \frac{-30}{(6x+6h + 3)(6x+3)}
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  3. #3
    Junior Member ImaCowOK's Avatar
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    That's exactly what I did, but how did you get rid of the h in the denominator? You got an error on one of your final steps.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    I missed a '}' in my post. I corrected it.

    Remember that:

    \frac{(\frac{a}{b})}{c} = \frac{a}{b} \times \frac{1}{c} = \frac{a}{bc}
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  5. #5
    Junior Member ImaCowOK's Avatar
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    Okay, I remember now. But I have to then use it to find \lim_{h \to 0}. I keep getting that wrong.

    The other parts like f'(2), and f'(3) I've gotten right. f'(1) apparently is wrong and I did it exactly the way I did the others.
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, now I'm not sure what you are talking about. Seems beyond what I've learned...

    But there is an 'h' in the function. As h tends to zero, 6h will tend to zero, and the function tends to:

        = \frac{-30}{(6x+ 3)(6x+3)}
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  7. #7
    Junior Member ImaCowOK's Avatar
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    f'(1) and such means the derivative of the function when x is 1, so plug 1 into the function wherever there is an x. Sorry if you haven't learned it yet and I confused you.

    Thanks for the answers you really helped me out. I had the right idea, but I was unsure about how to enter it.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Ok, I'm familiar with derivatives. It's the limit that I'm not familiar with.

    Uh... do you differentiate with the h? Because, if we already take it as h tends to zero, then,

    f(x) = \frac{-30}{(6x+ 3)(6x+3)} = -30(6x + 3)^{-2}

    f'(x) = 60(6x+3)^{-3} . 6 = \frac{360}{(6x+3)^3}
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  9. #9
    Junior Member ImaCowOK's Avatar
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    \lim_{h \to 0} was \frac{-30}{(6x+ 3)(6x+3)} just as you said.

    You substitute 0 in \frac{-30}{(6x+6h + 3)(6x+3)} for h which leaves you with \frac{-30}{(6x+ 3)(6x+3)} which was the correct answer. I didn't need to go any further for this particular exercise.
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