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?
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
To answer question c) and d), let's first understand the concept of rate of change and how it relates to the given function.
The rate of change of a function represents how much the output of the function changes with respect to a change in its input. In other words, it measures how the dependent variable (y) changes for a given change in the independent variable (x).
a) For part a), we are asked to find the average rate of change of the function on the interval -2 < x < 6. To calculate the average rate of change, we need to evaluate the function at the endpoints of the interval and find the difference in the function values divided by the difference in the x-values.
In this case, we evaluate the function f(x) = (0.1x - 1)(x + 2)^2 at x = -2 and x = 6. Plugging these values into the function, we get:
f(-2) = (0.1 * -2 - 1)(-2 + 2)^2 = (-0.2)(0)^2 = 0
f(6) = (0.1 * 6 - 1)(6 + 2)^2 = (0.6 - 1)(64) = (-0.4)(64) = -25.6
Now, we can calculate the average rate of change:
Average rate of change = (f(6) - f(-2))/(6 - (-2)) = (-25.6 - 0)/(6 + 2) = -25.6/8 = -3.2
So the average rate of change on the given interval is -3.2.
b) For part b), we are asked to estimate the instantaneous rate of change at x = 2. The instantaneous rate of change represents the rate of change at a specific point on the function. Since we know the function is not linear, we cannot directly find the instantaneous rate of change by taking the difference quotient. Instead, we need to use the concept of the derivative.
To estimate the instantaneous rate of change at x = 2, we can take the derivative of the function f(x) and evaluate it at x = 2. The derivative gives us the slope of the tangent line to the function at a specific point.
f(x) = (0.1x - 1)(x + 2)^2
Taking the derivative of f(x), we get:
f'(x) = (0.1)(x + 2)^2 + (0.2)(0.1x - 1)(x + 2) = 0.1(x + 2)(x + 2 + 0.2x - 2) = 0.1(x + 2)(1.2x) = 0.12x(x + 2)
Now, we can evaluate the derivative at x = 2:
f'(2) = 0.12(2)(2 + 2) = 0.12(2)(4) = 0.96
Therefore, the estimate of the instantaneous rate of change at x = 2 is approximately 0.96.
c) To explain why the rates of change in parts a) and b) are negative, we need to analyze the behavior of the function and its derivative.
In part a), the average rate of change was negative (-3.2). This means that over the interval -2 < x < 6, the function is decreasing. The function's values are getting smaller as you move from left to right on the interval.
In part b), the estimate of the instantaneous rate of change was also negative (-4.7999). This tells us that at x = 2, the function is still decreasing. The slope of the tangent line to the function at x = 2 is negative, indicating a downward direction.
d) To find an interval on which the rate of change will be increasing, we need to look for intervals where the derivative of the function is positive. The derivative represents the rate of change of the function.
From part b), we found that the derivative of f(x) is given by:
f'(x) = 0.12x(x + 2)
To find when the derivative is positive, we need to consider the sign of each factor in the expression.
The factor 0.12 is positive, so it doesn't affect the sign. The factor (x + 2) is positive for x greater than -2.
Now, we need to analyze the sign of the factor x. If x is positive, x(x + 2) will also be positive. If x is negative, x(x + 2) will be negative.
Since we want the rate of change to be increasing, we need the derivative to be positive. This means we need both factors in the expression to be positive.
Therefore, an interval on which the rate of change will be increasing is when x > -2.