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I got through a lot of this type of math question on my homework last week, but got this one wrong. Was hoping you could help explain to me the answer and how you got it.

Given: f(x)=2x/(5x^(2)-x)

Find the average rate of change of f(x) on the interval [1,4]

Thank you!

First, the expression for f(x) can be simplified by factoring an x in the denominator:

f(x) = 2x/( 5x^2 - x ) = 2x/( x(5x - 1) ) = 2/(5x - 1).

The average of a function, say g(x), over the interval [a,b] is defined by the integral

g_ave(a,b) = (1/(b-a))･∫g(x)dx with limits a -> b.

For your case, we take g(x) = df(x)/dx = derivative of f(x) = f'(x) = rate of change of f(x). If we make the substation f'(x) = g(x), then by the Fundamental Theorem of Calculus

g_ave = f'_ave = 1/(b-a){ f(b) - f(a) }.

Plugging in the limits b = 4 and a = 1, I get

f'_ave = (1/3){2/19 - 2/4)

which you could reduce further if you want.

Hope this helps.

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