let f(x)= 2 - |2x-1|. Show that there is no value of c such that f(3)- f(0) = f'(c)(3-0). Why does this not contradict the mean value theorem.
Yes
To determine if there is a value of c such that f(3) - f(0) = f'(c)(3 - 0), we need to evaluate each component of the equation and see if they are compatible.
1. Calculate f(3) - f(0):
Substitute x = 3 into f(x) and x = 0 into f(x) to get the values of f(3) and f(0):
f(3) = 2 - |2(3)-1| = 2 - |6-1| = 2 - 5 = -3
f(0) = 2 - |2(0)-1| = 2 - |0-1| = 2 - 1 = 1
Therefore, f(3) - f(0) = (-3) - 1 = -4.
2. Calculate f'(c):
To find f'(x), we need to determine the derivative of f(x) and set it equal to f'(c) to evaluate it at some c.
Derivative of f(x):
f(x) = 2 - |2x - 1|
To find the derivative, we need to consider two cases:
a) When 2x - 1 ≥ 0:
In this case, |2x - 1| = 2x - 1. So, f(x) = 2 - (2x - 1) = 2 - 2x + 1 = 3 - 2x.
The derivative of f(x) is:
f'(x) = -2
b) When 2x - 1 < 0:
In this case, |2x - 1| = -(2x - 1) = -2x + 1. So, f(x) = 2 - (-2x + 1) = 3 + 2x.
The derivative of f(x) is:
f'(x) = 2
Since f'(x) is not a constant function, it does not have a fixed value.
3. Find f'(c)(3 - 0):
Substitute f'(x) = -2 and 3 - 0 = 3 into the equation to get:
f'(c)(3 - 0) = (-2)(3) = -6
Conclusion:
We have f(3) - f(0) = -4 and f'(c)(3 - 0) = -6. Since -4 is not equal to -6, it means there is no value of c such that f(3) - f(0) = f'(c)(3 - 0).
However, this does not contradict the Mean Value Theorem. The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In this case, the function f(x) is not differentiable at x = 1 since the absolute value function is not differentiable at that point. Therefore, the Mean Value Theorem does not apply, and the lack of a value c that satisfies f(3) - f(0) = f'(c)(3 - 0) does not contradict the theorem.