1. ## 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.

2. \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):
\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:
\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:
\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}

3. 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 . . .

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

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

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

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

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

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

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

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