Question

How do I solve this math problem?

I need help can someone show me how to do this step by step?

Find the average rate of change with the given function.

g(x) = 1/x; x = 1, x = a

Answer

If you have learned about derivatives in calculus, then find the derivative (same as the rate of change) of (1/x), which is (-1/x^2). So that the rate of change at x = 1 is -1, and the rate of change at x = a is (-1/a^2).

So the average rate of change is

[(-1) + (-1/a^2)]/2. I will let you simplify this algebraically to get the final form of the answer.

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If you don't know about derivatives, then there is another way to handle this. Take a positive number, represented here by "y", and assume that it is very small compared to either 1 or the absolute value of "a", and in fact "y" can be chosen to be as small as desired (one millionth, one billionth, etc).

At any value which "x" can have, an approximation for the rate of change can be found by taking:

[1/(x+y) - 1/(x-y)]/[(x+y)-(x-y)] = [change in f(x)/change in x] = rate of change.

It should be easy to see that, since the rate of change is not the same at different values of "x", the smaller you make the "y" value in this equation, the more accurate will be the rate of change found for the slope at x = a, or for any other value of "x".

The expression on the left side of the equation can be manipulated to give:

{[(x-y) - (x+y)]/[ (x+y)(x-y)]}/(2x) = {-2y/(x^2-y^2)}/2y} = -1/(x^2-y^2)

No matter what size of number value we give to "x" in this last expression, we can always choose a "y" value so much smaller than that value of "x", it will be too small to matter at all. This means that the expression for the rate of change will be -1/(x^2) to as close an approximation as is needed or possible. This is where the calculus derivative above for 1/x comes from, so just use it the same way as above to find the AVERAGE rate of change over x = 1 to x = a by substitution of these values into the (-1/x^2) expression for the "instantaneous" rate of change given by the (-1/x^2).

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