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I am give $g'$ of a series of numbers and I am suppose to rank them in increasing order. I have no idea what this means. I think it meant to take that as an $x$ point on the graph and then find the value of of $f(x)$ but that was not correct at all.

I think what the question wants me to do is find the derivative. $g'(-2)$ so I am guessing I would take $-2$ and plug it into the $f(a+h)-f(a) / h$ somehow but I am not sure how. What is what in this problem? What is $a$ and what is $h$?

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17. For the function g whose graph is given , arrange the following numbers in increasing order and explain your reasoning: 0 g^1(-2) g^1(8) g^1(2) g^1(4) – user138246 Sep 11 '11 at 20:22
I am not sure how to do that, is only has x values on it. – user138246 Sep 11 '11 at 20:24
For instance, $f^{\prime }(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$, $f^{\prime }(b)=\lim_{h\rightarrow 0}\frac{f(b+h)-f(b)}{h}$, $f^{\prime }(c)=\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}$. Assume $f^{\prime }(a)\leq f^{\prime }(b)\leq f^{\prime }(c)$. Are you asked to order them in such a way? – Américo Tavares Sep 11 '11 at 20:26
So f^1(a) is just the limit of the average velocity formula? – user138246 Sep 11 '11 at 20:31
@Jordan: Please don't accept an answer until you are clear and satisfied with it; my previous answer to your questions was/is followed by many comments asking follow-up questions, even though you had already accepted the answer. First ask your follow-ups and make sure everything is clear, then accept if you are satisfied. Also, by not accepting, you may be inviting other people to provide answers you may find more helpful. In short: wait a bit and try to digest the answer thoroughly before accepting! – Arturo Magidin Sep 11 '11 at 20:38

1 Answer 1

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You are given a graph that you can look at. You are asked to rank $g'(-1)$, $g'(8)$, $g'(2)$, and $g'(4)$ in increasing order.

What is "$\mathbf{g'(a)}$"?

It's the slope of the tangent to the graph through the point $\mathbf{(a,g(a))}$.

So: $g'(-1)$ is the slope of the tangent to the graph you are looking at, through the point with $x$-coordinate $-1$. $g'(8)$ is the slope of the tangent to the graph you are looking at through the point with $x$-coordinate $8$. $g'(2)$ is the slope of the tangent through the point with $x$-coordinate $2$ of the graph you are looking at. $g'(4)$ is the slope of the tangent to the graph you are looking at, through the point with $x$-coordinate $4$.

But they are numbers, because the "slope of the tangent" is itself a number.

Since you can see the graph, you can estimate what the tangent will look like graphically. Since you can estimate what the tangent will look like, you can see which tangent is steepest (has the largest slope), and which tangent has the smallest slope. Use those estimates to order the numbers $g'(-1)$, $g'(8)$, $g'(2)$, and $g'(4)$ in order.

The number $0$ would represent a tangent line with slope $0$, that is, horizontal. Positive slopes are lines that "rise" as you move left to right; negative slopes are lines that "fall" as you move left to right. Again: since you can see the graph, and you can estimate what the tangents will be, you should be able to figure out which tangents have positive slope (and go after the $0$ in the list you will give), and which ones have negative slope (and go before $0$ in the list you will give).

In short: This question (a typical question from many a calculus book) is asking you to use what the derivative represents (not how it is computed) to describe at which points the function whose graph you are seeing has steepest or flattest graph, where it is "decreasing rapidly", where it is decreasing but slower, where it is increasing a bit, and where it is increasing the fastest, and to place $0$ to separate the points where the graph has negative slope and where the graph has positive slope.

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So if I am given f^1(4) all I know is that it is the value of the slope at 4, f(4)? That seems odd to me that to slope would always be the same value as x. – user138246 Sep 11 '11 at 20:41
@Jordan: From the phrasing of the question, you are given the graph of $f$ (or was it $g$?). You can look at the graph. You can see what the graph looks like at or around $x=4$. $f(4)$ is how high the graph is at $x=4$; $f'(4)$ is how steep the graph is at $x=4$. The slope is not $4$, the slope is some number that we call $f'(4)$. The sequence of symbols $f'(4)$ mean "the number that is the slope of the tangent to the graph of $y=f(x)$ at the point $x=4$". The slope is not "the same value as $x$". – Arturo Magidin Sep 11 '11 at 20:46
so the slope would be defined as f^1(4)=x? – user138246 Sep 11 '11 at 20:49
@Jordan: No. The slope is the slope of the tangent at the point. One way to compute it is with the limit of difference quotients, $$\lim_{x\to 4}\frac{f(x)-f(4)}{x-4}.$$But since you don't have a formula for $f$, you cannot use the formula here; instead, you have the graph of $f$, so what you can do is draw the tangent and then figure out the slope graphically. Whatever the slope is, that number is what "$f'(4)$" is equal to. $f'(4)$ is the name of the number, not the value. The value of $f'(4)$ is "the slope of the tangent to the graph at $x=4$". – Arturo Magidin Sep 11 '11 at 20:53

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