1. ## Calculus problem

A bicyclist is riding on a path modeled by the function f(x) = 0.04(8x -x^2), where x and f(x) are measured in miles. Find the rate of change of elevation when x = 2.

f(x) = 0.04(8x-x^2)

The solution manual I looking over says the answer is slope of tangent line at x = 2 is rate of change, and equals about 0.16.

I have no idea how this is about 0.16, so if anyone could shed some light it would be greatly appreciated.

2. Hi

The change of elevation is the slope of the tangent at x=2
In other words it is the value of the derivative at x=2

$\displaystyle f(x) = 0.04(8x-x^2)$

$\displaystyle f'(x) = 0.04(8-2x)$

$\displaystyle f'(2) = 0.04 \times 4 = 0.16$

3. y'=0.04(8-2x)
=0.04(4)
=0.16

4. We haven't learned derivatives in class yet, is there another way this problem can be solved.

5. You can use an approach like $\displaystyle \frac{f(2.01)-f(2)}{0.01}$

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# a bicyclist is riding on a path modeled by the function f(x)=.08x where x abd f(x) are measured in miles. Find the rate of change of elevation at x=2

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