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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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    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](http://math.stackexchange.com/posts/179436/revisions), and _then_ added an explanation of where in it you were stuck.2012-08-06

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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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    @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.2012-08-06
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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](http://math.stackexchange.com/posts/179436/revisions), 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.2012-08-06
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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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    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).2012-08-06