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Math Help - 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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    \begin{aligned}<br />
\frac{f(a + h) - f(a)}{h} &= \frac{\frac{15}{2a + 2h - 3} - \frac{15}{2a - 3}}{h}<br />
\end{aligned}

    The common denominator between 2a + 2h - 3 and 2a - 3 is their product, (2a + 2h - 3)(2a - 3):
    \begin{aligned}<br />
\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}<br />
\end{aligned}

    Multiply the two sets of fractions in the numerator:
    \begin{aligned}<br />
\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}<br />
\end{aligned}

    Now that the denominators are the same you can subtract the fractions:
    \begin{aligned}<br />
\frac{f(a + h) - f(a)}{h} &= \frac{  \frac{15(2a - 3) - 15(2a + 2h - 3)}{(2a + 2h - 3)(2a - 3)}}{h}<br />
\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: . f(x) \;=\;\dfrac{15}{2x-3}
    Newton's Quotient: . \dfrac{f(a+h) - f(a)}{h}

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

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

    . . 2. Subtract f(a) . . . find f(a) and subtract.

    . . 3. Divide by h.


    Here we go . . .


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


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

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

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

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

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


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

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