Math~Reinyyy
posted by Elli on .
Could someone help me with these questions, I don't know question c) and d)
Consider the function f(x) = (0.1x1)(x+2)^2.
a) Determine the function's average rate of change on 2<x<6.
Answer; Avg rate of change is 3.2
b) Estimate the instantaneous rate of change at x=2.
Answer 4.7999
c) Explain why the rates of change in parts a) and b) have been negative
d) Give an interval on which the rate of change will be increasing?

If the function is decreasing, the rate of change is negative.
The cubic function is zero at x = 2 and at x = 10
it is negative for large negative x
It is positive for large positive x
So it slopes down from x = 2 to x = +6
d) derivative = 2(.1x1)(x+2) + .1(x+2)^2
= 2[ .1 x^2 .8 x 2] +.1[x^2 + 4 x + 4]
= .2 x^2  1.6 x  4 +.1 x^2 +.4 x + .4
= .3 x^2  1.2 x  3.6
where is that zero?
0 = x^2  4 x  12
0 = (x6)(x+2)
so the slope of this function is zero at x = 2 and at x = 6
Sketch this function and you will see that the slope is only positive where x is less than 2 and where x is greater than 6