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(.1x-1)(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 = (x-6)(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
Could someone help me with these questions, I don't know question c) and d)
Consider the function f(x) = (0.1x-1)(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?
1 answer
1 answer