# Thread: help on a calculus problem

1. ## help on a calculus problem

find the change in elevation when X=2 for the path modeled by the function:
f(x)=0.04(8x-x^2), where x and f(x) are measured in miles...how can this problem be estimated using precalculus and how can it be solved with calculus?

any help would be appreciated thanks

2. Originally Posted by PMoNEY23
find the change in elevation when X=2 for the path modeled by the function:
f(x)=0.04(8x-x^2), where x and f(x) are measured in miles...how can this problem be estimated using precalculus and how can it be solved with calculus?

any help would be appreciated thanks
A rate of change is the slope of a curve at a point... which is the same as the slope of the line tangent to the curve at that point.

So you could draw a tangent to the curve at the point $x=2$ and measure its slope using $\frac{\textrm{Rise}}{\textrm{Run}}$.

You could also choose another point very close to 2, say 2.001, and work out the slope of the line that would cross the two points, using $\frac{\textrm{Rise}}{\textrm{Run}}=\frac{y_2-y_1}{x_2-x_1}$. This will give you a good approximation.

Using calculus, if a function $f(x)=ax^n$, its derivative (rate of change at any point) is given by $f'(x)=nax^{n-1}$. Substitute $x=2$ into the derivative and this will be the slope of the curve at that point.

3. how exactly can i find the derivative for that function? Can you plz explain in steps
Thanks

4. $f(x)=0.04(8x-x^2)$

Using the power rule that prove it mentioned

$f'(x)=nax^{n-1}$

with a being the coefficent and n the power

$f(x)= .32x - .04x^2$

$f'(x) = 32 - .08x$