What values (c) if any are predicted by the mean value theorem for the function f(x)= (x-2)^3 on the interval [0,2]?
I got x= 4 and x=0, since x=0 is within the interval I chose that as my answer. Thank you.
Sorry I forgot to include the topic.
MVT says that there is some c in [0,2] where f'(c) = [f(2)-f(0)]/2
f(2) = 0
f(0) = -8
[f(2)-f(0)]/2 = 8/2 = 4
MVT does not predict values. It just says that at some c in [0,2] f'(c) = 4.
Now, we can check that by noting
f'(x) = 3(x-2)^2, so
f'(0) = 8 and f'(2) = 0, so the slope decreases from 8 to 0 in the interval. Since f' is continuous, f'=4 somewhere on the interval.
3(x-2)^2 = 4
x-2 = ±2/√3
x = 2-2/√3 = 0.85
Thank you.
To determine the values (c) predicted by the mean value theorem for the function f(x) = (x-2)^3 on the interval [0,2], you need to check if the conditions of the mean value theorem are satisfied.
The mean value theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that the derivative of the function evaluated at c, f'(c), is equal to the average rate of change of the function over the interval [a, b], which is given by (f(b) - f(a))/(b - a).
In this case, the function f(x) = (x-2)^3 is continuous on the interval [0,2] and differentiable on the open interval (0,2). Therefore, the conditions required for the mean value theorem are satisfied.
Now, let's find the derivative of f(x) to determine the average rate of change over the interval [0,2]:
f'(x) = 3(x-2)^2
To find the average rate of change, we evaluate f'(x) at the endpoints of the interval [0,2] and calculate the difference:
f'(2) - f'(0) = 3(2-2)^2 - 3(0-2)^2
= 0 - 3(-2)^2
= 0 - 3(4)
= -12
Since the average rate of change over the interval [0,2] is -12, according to the mean value theorem, there exists at least one value c in (0,2) such that f'(c) = -12.
To find this value, we need to solve f'(c) = -12:
3(c-2)^2 = -12
Simplifying the equation:
(c-2)^2 = -4
The equation (c-2)^2 = -4 has no real solutions since a square of a real number cannot be negative. Therefore, there is no value (c) predicted by the mean value theorem for the function f(x) = (x-2)^3 on the interval [0,2].
Hence, your answer of x = 0 is incorrect; there is no value of c within the interval [0,2] predicted by the mean value theorem for this function.