graphing and functions

one of my problem says that the average slope formula is: $\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$ It tells us the average slope over the interval from $x_1$ to $x_2$ Then it says to take the average slope of the function $x^2+3x+1$ over one unit intervals from $0$ to $10$.

can i use a particular function function to solve this problem?

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You did not give the exact wording of the question, but it sounds like they want $10$ different numbers, $(f(1)-f(0))/(1-0)$, $(f(2)-f(1))/(2-1)$, and so on. – André Nicolas Aug 6 '12 at 13:09
$f(x)=x^2+3x+1, x_1=0, x_2=10$ – i. m. soloveichik Aug 6 '12 at 13:14
This question was better before it was edited. And it would have been even better if you had quoted the entire exercise like you did originally, and then added an explanation of where in it you were stuck. – Henning Makholm Aug 6 '12 at 13:43

As André said, what they want is the 10 numbers $$\frac{f(1)-f(0)}{1-0}, \frac{f(2)-f(1)}{2-1}, \ldots, \frac{f(10)-f(9)}{10-9}$$ -- that is, with $x_1$ increasing from $0$ to $9$ while $x_2$ increases from $1$ to $10$. Then plot them against the $x_1$ values for each of them, and look for a pattern.

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so do i just do 1-0/1-0+2-1/2-1...10-9/10-9 and then plug the answer to the equation x2+3x+1? – Aleef Ahmed Aug 6 '12 at 19:07
@AleefAhmed: No! $x^2+3x+1$ is the definition of $f(x)$, so when you need to compute $f(0)$, you plug $0$ into that formula; when you need to compute $f(1)$ you plug $1$ into that formula, and so forth. This happens twice for each of the slopes you compute (though you can save yourself some time by remembering what, say, $f(1)$ is, rather than computing it twice). Finally, who says you should add the results? You shouldn't -- the ten different results are what you should plot as 10 different points on your graph paper. – Henning Makholm Aug 6 '12 at 19:24
so the 10 different results are my x value? what do i do with the definition of f(x) – Aleef Ahmed Aug 6 '12 at 19:41
@AleefAhmed: $f(x)=x^2+3x+1$. The question seems to be asking for the $10$ quantities $\frac{f(1)-f(0)}{1-0}$,$\frac{f(2)-f(1)}{2-1}$,$\frac{f(3)-f(2)}{3-2}$, $\dots$, $\frac{f(9)-f(10)}{10-9}$. Nothing fancier than that. – robjohn Aug 6 '12 at 20:10

Since each interval is of length $1$ you could look at $$\dfrac{f(x+1)-f(x)}{(x+1)-(x)} = \dfrac{((x+1)^2+3(x+1)+1)-(x^2+3x+1) }{(x+1)-(x)}$$ for $x=0,1,2,\ldots,9$. You might find it easier if you simplified the expression.

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In light of Aleef's earlier attempt to ask the question, I don't think symbolic simplification is the way to go here. Just evaluate $f(0)$, $f(1)$, ..., $f(10)$ using the original definition of $f$, and then plug those values as numbers into the given average slope formula. – Henning Makholm Aug 6 '12 at 13:19

The naive approach is to make yourself a table, with $10$ rows (one for each $x_i$ with: $x_1 = 1, x_2 = 2, \dots, x_{10} = 10$). The first column would be $f(x_i)$, the second column would be $x_i - x_{i-1}$ (you should notice an obvious pattern in this column), the third column would be $f(x_i) - f(x_{i-1})$, and the fourth column would be:

$\dfrac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}}$

(and if you noticed the pattern in the second column, you might not need this last one....why?)

To get this started, set $x_0 = 0$, with $f(x_0) = f(0) = 1$.

Note that Henry's answer above gives a short-cut, this is "the long way".

The first row (for $x_1 = 1$) would look like this:

$f(1) = 5;\ \ x_1 - x_0 = 1 -0 = 1;\ \ f(1) - f(0) = 5-1 = 4;$ slope $= 4/1 = 4$.

Nine more to go....

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$f(x)=x^2+3x+1, x_1=0, x_2=10$ so the average slope on the interval [0,10] is $\frac{f(10)-f(0)}{10-0}=\frac{131-1}{10}=13$.

If you want the average slope over all 1 unit intervals $[a, a+1]$ for $0\le a\le 9$ then this is the symbolic expression $\frac{f(a+1)-f(a)}{a+1-a}$ which simplifies to $2a+4$.

For example with $a=4.5$ we also get average slope of 13, but for $a=9$ we get average slope of 22.

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No -- the exercise asks for average slopes over one-unit intervals from 0 to 10, not over a single 10-unit interval. – Henning Makholm Aug 6 '12 at 13:21
This still answers the wrong question. The average slope over the entire interval is not being asked for. What the question is asked for is the average slope over each one-unit interval, separately. (Though, perhaps, that may only be apparent if one has read the OP's earlier, now deleted, question). – Henning Makholm Aug 6 '12 at 13:39