Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of c in that interval that satisfy the conclusion of the theorem.
f(x)=x^2-3x; [-2,6]
To verify whether the hypotheses of the Mean Value Theorem (MVT) are satisfied for the function f(x) = x^2 - 3x on the interval [-2, 6], we need to check the following conditions:
1. f(x) should be continuous on the closed interval [-2, 6].
2. f(x) should be differentiable on the open interval (-2, 6).
Now, let's check each condition:
1. Continuity: The function f(x) = x^2 - 3x is a polynomial, which is continuous for all real values of x. Therefore, it is continuous on the closed interval [-2, 6].
2. Differentiability: To check differentiability, we need to determine whether f(x) is differentiable for all values of x in the open interval (-2, 6).
The function f(x) = x^2 - 3x is a polynomial, and all polynomials are differentiable for all real values of x. Therefore, it is differentiable on the open interval (-2, 6).
Now, as we have verified that f(x) is both continuous and differentiable on the interval [-2, 6], we can conclude that the hypotheses of the Mean Value Theorem are satisfied.
Next, we need to find all the values of c in the interval (-2, 6) that satisfy the conclusion of the Mean Value Theorem.
According to the Mean Value Theorem, there exists at least one value c in the open interval (-2, 6) such that the derivative of f(x) at c is equal to the average rate of change of f(x) over the interval [-2, 6].
The average rate of change of f(x) over the interval [-2, 6] is given by:
( f(6) - f(-2) ) / ( 6 - (-2) )
Now, let's calculate the values:
f(6) = (6^2) - 3(6) = 36 - 18 = 18
f(-2) = ((-2)^2) - 3(-2) = 4 + 6 = 10
So, the average rate of change is:
( f(6) - f(-2) ) / ( 6 - (-2) ) = (18 - 10) / 8 = 8 / 8 = 1
Now, we need to find the value of c in the interval (-2, 6) that satisfies:
f'(c) = 1
To find this value, we will take the derivative of f(x) and set it equal to 1:
f'(x) = d/dx(x^2 - 3x) = 2x - 3
Equating f'(x) with 1:
2x - 3 = 1
Solving for x:
2x = 4
x = 2
Therefore, the value of c in the interval (-2, 6) that satisfies f'(c) = 1 is c = 2.
To summarize, the hypotheses of the Mean Value Theorem are satisfied on the interval [-2, 6] for the function f(x) = x^2 - 3x, and the value of c in the interval (-2, 6) that satisfies f'(c) = 1 is c = 2.