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Evaluate and simplify different quotients $\frac{f(x+h)-f(x)}{h}$ where $h\neq 0$ and $\frac{f(x)-f(a)}{x-a}$ where $x\neq a$ for each of the following functions.

a. $f(x)=x^2-2x$

b. $f(x)=4-5x$

c. $f(x)=x^3+2$

d. $f(x)=\frac{7}{x+3}$

Hi, all. I tried looking through my notes for these exercises, but unfortunately, I cannot find them. Can someone please explain how to do at least one of these questions? Thank you! I greatly appreciate any advice/solutions.

NOTE: Also, I kind of forgot what those quotient functions things are. Can someone please explain to me what in the world are they? Thanks again! :)

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  • $\begingroup$ Just a clarification: when I look at this question, I'm really confused. What's $h$? $\endgroup$ – 关一骏 Aug 17 '16 at 23:25
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    $\begingroup$ $h$ is just a perturbation of $x$ $\endgroup$ – Hamza Aug 17 '16 at 23:27
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    $\begingroup$ There are tons of similar questions under "related" on the right side of your screen. Also you can search: youtube.com/results?search_query=difference+quotient . There are tons of resources online, just use Google. $\endgroup$ – mathematician Aug 17 '16 at 23:27
  • $\begingroup$ @mathematician, oh thanks! $\endgroup$ – 关一骏 Aug 17 '16 at 23:28
  • $\begingroup$ Problem: Google is not loading so RIP. $\endgroup$ – 关一骏 Aug 17 '16 at 23:28
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I will do b) as an example. $$f(x)=4-5x$$ gives \begin{align} \frac{f(x+h)-f(x)}{h} &= \frac{4-5(x+h)-(4-5x)}{h} \\ &= \frac{4-5x+5x-5h-4}{h} \\ &= \frac{-5h}{h} \\ &= -5. \end{align}

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as example for $a)$ \begin{eqnarray} \frac{f(x+h)-f(x)}{h}&=& \frac{(x+h)^2-2(x+h)-(x^2-2x)}{h}\\ &=& \frac{x^2+h^2+2xh-2x-2h-x^2+2x}{h}\\ &=& \frac{h^2+2xh-2h}{h}\\ &=& \frac{2xh+h^2}{h}-\frac{2h}{h}\\ &=& 2x+h-2 \end{eqnarray}

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  • $\begingroup$ The final answer is in that form? Oh, I thought that it had to be an actual number! Thanks! $\endgroup$ – 关一骏 Aug 17 '16 at 23:34
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    $\begingroup$ you see that we can't develop the last line, but if we take $h$ to zero we will get the derivative of $f$ where we don't see any $h$ in the formulation $\endgroup$ – Hamza Aug 17 '16 at 23:36
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    $\begingroup$ Everything's starting to come back to me—starting to remember this stuff again! Thanks! $\endgroup$ – 关一骏 Aug 17 '16 at 23:38
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    $\begingroup$ you are welcome $\endgroup$ – Hamza Aug 17 '16 at 23:44
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Just insert the specific $f(x)$ in $\frac{f(x+h)-f(x)}{h}$:

b) gives $$\frac{4-5(x+h)-(4-5x)}{h}=\frac{-5h}{h}=-5$$

EDIT: As per OP's request, these things are called "difference quotients" and when you take the limit $h\rightarrow 0$ or (in the other case) $x\rightarrow a$, you get the derivative $\frac{df}{dx}$.

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    $\begingroup$ picked the linear example as well i see :p $\endgroup$ – qbert Aug 17 '16 at 23:31
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    $\begingroup$ @qbert Damn, you beat me to it! :) (+1) $\endgroup$ – Bobson Dugnutt Aug 17 '16 at 23:32
  • $\begingroup$ OH, THAT'S WHAT IT MEANS? I got so confused by the other questions. Thank you so much! $\endgroup$ – 关一骏 Aug 17 '16 at 23:32
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    $\begingroup$ just barely haha $\endgroup$ – qbert Aug 17 '16 at 23:32
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  • $c)$

\begin{align} \frac{f(x+h)-f(x)}{h} &= \frac{(x+h)^3 +2 -x^3 -2}{h} \\ &= \frac{3x^2h +3xh^2 +h^3}{h} \\ &= 3x^2 +3xh +h^2 \end{align}

  • $d)$

\begin{align} \frac{f(x+h)-f(x)}{h} &= \frac{\dfrac{7}{x+h+3} -\dfrac{7}{x+3}}{h} \\ &= \frac{7(x+3-x-h-3)}{h(x+3)(x+h+3)} \\ &= -\frac{7}{(x+3)(x+h+3)} \end{align}

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  • $\begingroup$ I think you made a typo! :P $\endgroup$ – 关一骏 Aug 17 '16 at 23:39
  • $\begingroup$ where exactly?? $\endgroup$ – haqnatural Aug 17 '16 at 23:41
  • $\begingroup$ I think you fixed it! d) to c)! $\endgroup$ – 关一骏 Aug 17 '16 at 23:43
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No-ones showing the second quotient:

a) $f(x) = x^2 - 2x$

So

\begin{align} \frac{f(x+h) - f(x)}{h} &= \frac{(x +h)^2 - 2(x-h) - x^2 + 2x}{h} \\ &= \frac{x^2 + 2hx + h^2 - 2x - 2h - x^2 + 2x}h \\ &= \frac{2hx + h^2 - 2h}h \\ &= 2x +h - 2 \end{align}

And for the other

\begin{align} \frac{f(x) -f(a)}{x-a} &= \frac{x^2 -2x -a^2 +2a}{x-a} \\ &= \frac{(x^2 -a^2) - (2x -2a)}{x-a} \\ &= \frac{(x+a)(x-a) - 2(x-a)}{x-a} \\ &= \frac{((x+a) -2)(x-a)}{x-a} \\ &= x +a -2 \end{align}

Notice if you replace $a$ with $x + h$, get the same thing.

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(This is all hinting toward learning about derivatives in calculus by the way.)

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

and

$$\lim_{a\rightarrow x}\frac{f(x) - f(a)}{h} = \lim_{a\rightarrow x} (x +a - 2) = x + x -2= 2x -2$$

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If you think of what these equations are doing graphically$\ldots$

They are taking two points, dividing the vertical change in value by the horizontal change in value. In other words, they are taking the slope of the line between these points.

What happens if the points get very close together? If $h$ becomes tiny or $x -a$ becomes tiny? Then this is the slope of two points on the graph very close together. As $h$ approaches zero or $a$ approaches $x$ this will become very close to the slope of the tangent line at $x$ of the function.

That's why these quotients are a big deal.

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  • $\begingroup$ Wow, thanks! @fleablood $\endgroup$ – 关一骏 Aug 18 '16 at 0:24

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