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Thread: Functions: Newton's Quotient

  1. #1
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    Functions: Newton's Quotient

    I'm doing some math over the summer for fun. I have a very weak background in math, but have come around and thoroughly enjoy it. Can someone explain what is going on with this problem involving Newton's Quotient?

    I was doing fine, until I hit some a algebra concept I'm unfamiliar with. It's the third photo were my pencil is pointing to. The numerator suddenly become 15(2a-3)-15(2a+2h-3). I don't understand how that is.






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  2. #2
    Senior Member eumyang's Avatar
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    $\displaystyle \begin{aligned}
    \frac{f(a + h) - f(a)}{h} &= \frac{\frac{15}{2a + 2h - 3} - \frac{15}{2a - 3}}{h}
    \end{aligned}$

    The common denominator between 2a + 2h - 3 and 2a - 3 is their product, (2a + 2h - 3)(2a - 3):
    $\displaystyle \begin{aligned}
    \frac{f(a + h) - f(a)}{h} &= \frac{ \left( \frac{15}{2a + 2h - 3} \right) \left( \frac{2a - 3}{2a - 3} \right) - \left( \frac{15}{2a - 3} \right) \left( \frac{2a + 2h - 3}{2a + 2h - 3} \right) }{h}
    \end{aligned}$

    Multiply the two sets of fractions in the numerator:
    $\displaystyle \begin{aligned}
    \frac{f(a + h) - f(a)}{h} &= \frac{ \frac{15(2a - 3)}{(2a + 2h - 3)(2a - 3)} - \frac{15(2a + 2h - 3)}{(2a - 3)(2a + 2h - 3)}}{h}
    \end{aligned}$

    Now that the denominators are the same you can subtract the fractions:
    $\displaystyle \begin{aligned}
    \frac{f(a + h) - f(a)}{h} &= \frac{ \frac{15(2a - 3) - 15(2a + 2h - 3)}{(2a + 2h - 3)(2a - 3)}}{h}
    \end{aligned}$
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  3. #3
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    Hello, MathIsCreative!

    I'm doing some math over the summer for fun. . Good for you!

    5, Evaluate Newton's Quotient for: .$\displaystyle f(x) \;=\;\dfrac{15}{2x-3}$
    Newton's Quotient: .$\displaystyle \dfrac{f(a+h) - f(a)}{h}$

    I like to break up Newton's Quotient into three steps:

    . . 1. Find $\displaystyle f(a+h)$ . . . replace $\displaystyle x$ with $\displaystyle a+h.$

    . . 2. Subtract $\displaystyle f(a)$ . . . find $\displaystyle f(a)$ and subtract.

    . . 3. Divide by $\displaystyle h.$


    Here we go . . .


    $\displaystyle 1.\;\;f(a+h) \:=\:\dfrac{15}{2(a+h) - 3} \;=\;\dfrac{15}{2a+2h-3}$


    $\displaystyle 2.\;\;f(a+h) - f(a) \;=\;\dfrac{15}{2a+2h-3} - \dfrac{15}{2a-3} $

    . . . . . . . . . . . . . .$\displaystyle =\;\dfrac{15}{2a+2h-3}\cdot\dfrac{2a-3}{2a-3} - \dfrac{15}{2a-3}\cdot\dfrac{2a+2h-3}{2a+2h-3} $

    . . . . . . . . . . . . . .$\displaystyle =\;\dfrac{15(2a-3) - 15(2a+2h-3)}{(2a+2h-3)(2a-3)}$

    . . . . . . . . . . . . . .$\displaystyle =\; \dfrac{30a - 45 - 30a - 30h + 45}{(2a+2h-3)(2a-3)}$

    . . . . . . . . . . . . . .$\displaystyle =\; \dfrac{-30h}{(2a+2h-3)(2a-3)}$


    $\displaystyle 3.\;\;\dfrac{f(a+h) - f(a)}{h} \;=\;\dfrac{-30h}{h\cdot(2a+2h-3)(2a-3)} $

    . . . . . . . . . . . . . . .$\displaystyle =\;\dfrac{-30}{(2a+2h-3)(2a-3)}$
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