f(x) = .1(x-10)(x+2)^2
So, it has a double root at x = -2 and another root at x = 10.
Since it's a cubic, with positive coefficient for x^3, it rises from the left, brushes against the x-axis at x = -2, decreases and comes back to cross again at x=10.
f is decreasing at the places of interest.
Any interval where x > 10 will have f increasing. Since you ask about instantaneous changes, I assume you know about derivatives.
f' = .1 ( 3x^2 - 12x - 36 )
f' = 0 at x = -2 and 6
So, f is increasing for x < -2 and x > 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