Verify that f(x) = x^3 − 2x + 6 satisfies the hypothesis of the Mean-Value Theorem over the interval [-2, 3]
and find all values of C that satisfy the conclusion of the theorem.
To verify if the function f(x) = x^3 − 2x + 6 satisfies the hypothesis of the Mean-Value Theorem over the interval [-2, 3], we need to check two conditions:
1. Continuity: The function f(x) must be continuous on the closed interval [-2, 3].
2. Differentiability: The function f(x) must be differentiable on the open interval (-2, 3).
Let's check these conditions step-by-step:
1. Continuity:
To check for continuity, we need to make sure that f(x) is continuous for all values of x in the interval [-2, 3].
Since f(x) = x^3 − 2x + 6 is a polynomial, it is continuous for all real numbers. Hence, f(x) satisfies the continuity condition.
2. Differentiability:
To check for differentiability, we need to make sure that f(x) is differentiable for all values of x in the open interval (-2, 3).
Since f(x) = x^3 − 2x + 6 is a polynomial, it is differentiable for all real numbers. Hence, f(x) satisfies the differentiability condition.
Therefore, f(x) = x^3 − 2x + 6 satisfies the hypothesis of the Mean-Value Theoremover the interval [-2, 3].
Next, let's find the values of C that satisfy the conclusion of the theorem.
According to the Mean-Value Theorem, if f(x) satisfies its hypothesis, then there exists at least one value c in the interval (-2, 3) such that the derivative of f(x) at c is equal to the average rate of change of f(x) over the interval [-2, 3].
The average rate of change of f(x) over the interval [-2, 3] can be calculated as:
Average rate of change = (f(3) - f(-2))/(3 - (-2))
Let's calculate the average rate of change:
f(3) = (3)^3 - 2(3) + 6 = 27 - 6 + 6 = 27
f(-2) = (-2)^3 - 2(-2) + 6 = -8 + 4 + 6 = 2
Average rate of change = (27 - 2)/(3 - (-2)) = 25/5 = 5
Now, we need to find the value of c in (-2, 3) such that the derivative of f(x) at c equals 5.
The derivative of f(x) is given by:
f'(x) = 3x^2 - 2
Let's set f'(c) = 5 and solve for c:
3c^2 - 2 = 5
3c^2 = 7
c^2 = 7/3
c = ±√(7/3)
Hence, the values of c that satisfy the conclusion of the Mean-Value Theorem are c = √(7/3) and c = -√(7/3).
To verify if the function f(x) = x^3 - 2x + 6 satisfies the hypothesis of the Mean Value Theorem (MVT) over the interval [-2, 3], we need to check two conditions:
1. f(x) is continuous over the interval [-2, 3].
2. f(x) is differentiable over the open interval (-2, 3).
First, let's check for continuity:
To determine if f(x) is continuous over the interval [-2, 3], we need to check if it is continuous at each endpoint and all the points in between.
At x = -2, substituting the value into the function:
f(-2) = (-2)^3 - 2(-2) + 6 = -8 + 4 + 6 = 2
Since f(-2) exists and is a real number, the function is continuous at x = -2.
At x = 3, substituting the value into the function:
f(3) = (3)^3 - 2(3) + 6 = 27 - 6 + 6 = 27
Since f(3) exists and is a real number, the function is continuous at x = 3.
Now let's check for differentiability:
To determine if f(x) is differentiable over the interval (-2, 3), we need to find the derivative of the function and check if it exists for all values of x in the open interval (-2, 3).
The derivative of f(x) = x^3 - 2x + 6 is given by:
f'(x) = 3x^2 - 2
The derivative exists for all real numbers, so it exists for all values of x in the open interval (-2, 3).
Since f(x) is continuous over the interval [-2, 3] and differentiable over the open interval (-2, 3), it satisfies the hypothesis of the Mean Value Theorem.
Now, to find the values of c that satisfy the conclusion of the theorem, we need to find a value c in the open interval (-2, 3) such that:
f'(c) = (f(3) - f(-2)) / (3 - (-2))
Substituting the values into the equation:
3c^2 - 2 = (27 - 2) / (3 + 2)
3c^2 - 2 = 25 / 5
3c^2 - 2 = 5
3c^2 = 7
c^2 = 7/3
c = ±√(7/3)
So, the values of c that satisfy the conclusion of the theorem are c = √(7/3) and c = -√(7/3).