rate of change f(x) = 0.04(8x-x^2) x=2

[kpx001]

how would i solve this?
f(x) = 0.04(8x-x^2) to find the avg rate of change of elevation when x=2

i tried avg rate of change formula using values 2,3 (i do not know what 2nd value to use) and got .48 = a .6 = b when i plugged it in.
so .6-.48/3-2 = .12 ?

 

[Deleted member 4993]

how would i solve this?
f(x) = 0.04(8x-x^2) to find the avg rate of change of elevation when x=2

i tried avg rate of change formula using values 2,3 (i do not know what 2nd value to use) and got .48 = a .6 = b when i plugged it in.
so .6-.48/3-2 = .12 ?

Sometimes - the instantaneous rate is defined as the average rate at an instant ( I don't agree with that definition - but it is there).

How did your text book define "instantaneous rate"?

As posted, the question-author is asking for instantaneous rate.

 

[kpx001]

well i dont use calculus, but pre-cal if i can to get the answer, to get a numerical approach to estimate solution

 

[kpx001]

theres a graph and it is a word problem. "A bicyclist is riding on a path modeled by the function f(x) = .04(8x-x^2) where x and f(x) are measured in miles. Find the rate of change of elevation when x=2

 

[stapel]

i tried avg rate of change formula

What formula did they give you? Please define the terms precisely.

Thank you!

Eliz.

 

[kpx001]

they didn't give me a formula but the answer is .16 but i dont know how they got the answer

 

[skeeter]

.16 is the the instantaneous rate of change of elevation at x = 2 miles.

you need to determine the following limit to get that answer ...

\(\displaystyle \L \lim_{x \rightarrow 2} \frac{f(x) - f(2)}{x - 2}\)

 

[kpx001]

alright so
f(x) = 0.04(8x-x^2)
f(2) = .04(8(2)-2^2) = .48
f(3)=.04(8(3)-(3)^2) = .6

f(x)-f(2)/(x-2)
f(3)-f(2)/(3-2) = .6 - .48 / 1 = .12

or if its limits, i barely remember it.

 

[skeeter]

\(\displaystyle \L \lim_{x \rightarrow 2} \frac{f(x) - f(2)}{x - 2}\)

\(\displaystyle \L \lim_{x \rightarrow 2} \frac{.04(8x-x^2) - .48}{x - 2}\)

\(\displaystyle \L \lim_{x \rightarrow 2} \frac{.04(8x - x^2) - .04(12)}{x - 2}\)

\(\displaystyle \L \lim_{x \rightarrow 2} \frac{-.04(x^2 - 8x + 12)}{x - 2}\)

\(\displaystyle \L \lim_{x \rightarrow 2} \frac{-.04(x-2)(x-6)}{x - 2}\)

\(\displaystyle \L \lim_{x \rightarrow 2} -.04(x-6) = .16\)

 

[stapel]

they didn't give me a formula...

Okay then, what was the "average rate of change formula" that you did use? And what alternate method was presented in class and/or in your book? (Sometimes perfectly-valid solutions can be counted off if "wrong" methods were used, so it might be wise to use whatever they gave you in class.)

Thank you!

Eliz.