consider the function f(x)=x^3 - x^2 - 3x -2. find the average slope of this function on the interval (-2,3). by the mean value theorem, we know there exists a c in the open interval (-2,3) such that f'(c) is equal to this mean slope. find the two values of c in the interval which work.
To find the average slope of the function f(x) = x^3 - x^2 - 3x - 2 on the interval (-2, 3), we first need to find the derivative of the function.
Step 1: Find the derivative of f(x):
f(x) = x^3 - x^2 - 3x - 2
Using the power rule, we differentiate each term:
f'(x) = 3x^2 - 2x - 3
Step 2: Find the mean slope:
The average slope of a function on an interval (a, b) is given by the formula:
mean slope = (f(b) - f(a)) / (b - a)
In this case, a = -2 and b = 3, so we calculate the mean slope as follows:
mean slope = (f(3) - f(-2)) / (3 - (-2))
Step 3: Calculate f(3) and f(-2):
f(x) = x^3 - x^2 - 3x - 2
f(3) = (3)^3 - (3)^2 - 3(3) - 2
= 27 - 9 - 9 - 2
= 7
f(-2) = (-2)^3 - (-2)^2 - 3(-2) - 2
= -8 - 4 + 6 - 2
= -8
So, the mean slope is:
mean slope = (7 - (-8)) / (3 - (-2))
= 15 / 5
= 3
Step 4: Use the mean value theorem to find the values of c:
The mean value theorem states that there exists a c in the open interval (a, b) such that f'(c) = mean slope.
We need to find the values of c for which f'(c) = 3.
Set f'(c) = 3 and solve for c:
3x^2 - 2x - 3 = 3
Rearranging the equation:
3x^2 - 2x - 6 = 0
Factoring the quadratic equation:
(3x + 2)(x - 1) = 0
Setting each factor equal to zero:
3x + 2 = 0 or x - 1 = 0
Solving for x:
3x = -2 or x = 1
Dividing by 3:
x = -2/3
Therefore, there are two values for c in the interval (-2, 3) which work:
c = -2/3 and c = 1.
To find the average slope of the function on the interval (-2, 3), we need to calculate the average rate of change of the function over that interval. The average rate of change can be determined using the formula:
Average Slope = (f(b) - f(a)) / (b - a)
where "a" and "b" are the endpoints of the interval, in this case, -2 and 3.
So, plugging in the values into the formula, we have:
Average Slope = (f(3) - f(-2)) / (3 - (-2))
To find the values of f(3) and f(-2), we substitute the respective x-values into the function f(x):
f(3) = (3^3) - (3^2) - 3(3) - 2 = 27 - 9 - 9 - 2 = 7
f(-2) = (-2)^3 - (-2)^2 - 3(-2) - 2 = -8 - 4 + 6 - 2 = -8
Substituting these values into the formula, we have:
Average Slope = (7 - (-8)) / (3 - (-2)) = 15 / 5 = 3
So, the average slope of the function on the interval (-2, 3) is 3.
According to the mean value theorem, there exists at least one point "c" within the interval (-2, 3) for which the instantaneous rate of change (derivative) of the function f(x) is equal to the average rate of change (average slope) we just found.
To find the values of "c", we need to find the derivative of the function first.
Taking the derivative of f(x) = x^3 - x^2 - 3x - 2, we get:
f'(x) = 3x^2 - 2x - 3
Now, we need to find the points "c" within the interval (-2, 3) where f'(c) = 3.
Setting f'(x) equal to 3, we have:
3x^2 - 2x - 3 = 3
Rearranging the equation, we get:
3x^2 - 2x - 6 = 0
Now, we can solve this quadratic equation to find the values of "c".
Using factoring or the quadratic formula, we find that the equation has two solutions: x = -1 and x = 2.
Therefore, the two values of "c" within the interval (-2, 3) for which f'(c) = 3 are -1 and 2.
f(-2) = -8
f(3) = 7
average slope = (7+8)/(3+2) = 3
f'(x) = 3x^2 - 2x - 3
f'(c) = 3c^2 - 2c - 3
3c^2 - 2c - 3 = 3
3c^2 - 2c - 6 = 0
c = (2 ± √73)/6
= (2 ± 6√2)/6
= 1/3± √2