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?
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.
4 answers
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
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?
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?