Question 988799
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The average rate of change of a function is defined as the slope of the secant line between two points of interest on the graph of the function.


If you select a point *[tex \Large \left(t,f(t)\right)] on the function and then choose an amount of change in the independent variable of *[tex \Large h], then the new point will be *[tex \Large \left(t+h,f(t+h)\right)].  The average rate of change will be the slope of the line that passes through these two points.


Recall the formula for the slope of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


Compare this to the average rate of change formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{\Delta}{y}}{\text{\Delta}{x}}\ =\ \frac{f(x\ +\ h)\ -\ f(x)}{(x\ +\ h)\ -\ x}\ =\ \frac{f(x\ +\ h)\ -\ f(x)}{h}]


See illustration

 *[illustration AverageRateOfChange.jpg].


For your given function, just plug into the formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\frac{7}{t\ +\ h}\ -\ \frac{7}{t}}{h}\ =\ \frac{\frac{7t\ -\ 7(t\ +\ h)}{t^2\ +\ th}}{h}\ =\ \frac{-7}{t^2\ +\ th}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \