Question
How would I start this?
The amount that workers contribute monthly for health insurance premiums can be modeled by
A(t)=0.07t^3-3.1t^2+54.3t-230, where A is the monthly amount contributed and t is the number of years after 1980. Find the instaneous rate of change in monthly contribution in 2000.
The amount that workers contribute monthly for health insurance premiums can be modeled by
A(t)=0.07t^3-3.1t^2+54.3t-230, where A is the monthly amount contributed and t is the number of years after 1980. Find the instaneous rate of change in monthly contribution in 2000.
Answers
jim
Don't let all the digits and decimals throw you off.
The number of years after 1980 just means that in 1981, t = 1, and 1982, t = 2, and so on.
You need to differentiate
0.07t^3 - 3.1t^2 + 54.3t - 230
Which, really, is just differentiating
t^3, t^2, and t
and multiplying them by their respective constants.
Then find the valuw of the derivative function at t=20.
Is that enough of a start?
The number of years after 1980 just means that in 1981, t = 1, and 1982, t = 2, and so on.
You need to differentiate
0.07t^3 - 3.1t^2 + 54.3t - 230
Which, really, is just differentiating
t^3, t^2, and t
and multiplying them by their respective constants.
Then find the valuw of the derivative function at t=20.
Is that enough of a start?
Marilyn
I'm using the formula for instantaneous rate of change, but I'm getting -38.4 as my answer..what am I doing wrong?
560-1240+1086-230-560-1240+1086-230/20
-2480+2172-460/20
560-1240+1086-230-560-1240+1086-230/20
-2480+2172-460/20
bobpursley
A(t)=0.07t^3-3.1t^2+54.3t-230
A'=.21t^2-6.2t+ 54.3 Now you are a t=20 so
A'=84-12.4+54.3
How did you get -38?
A'=.21t^2-6.2t+ 54.3 Now you are a t=20 so
A'=84-12.4+54.3
How did you get -38?
Marilyn
If the question says use the formal definition of a derivative aren't I supposed to use f(x)=f(x+h)-f(x)/h?