Find a point π satisfying the conclusion of the Mean Value Theorem for the function π(π₯)=π₯^(β1) on the interval [1,3]
c=
To find a point c satisfying the conclusion of the Mean Value Theorem for the function f(x) = x^(-1) on the interval [1, 3], we need to first determine if the function satisfies all the conditions required by the Mean Value Theorem.
1. Continuity: The function f(x) = x^(-1) is continuous on the interval [1, 3] because it is a power function and is defined for all x in the interval.
2. Differentiability: The function f(x) = x^(-1) is differentiable on the interval (1, 3) as it is differentiable for all x in the open interval due to its algebraic expression.
Now that we have established the prerequisites for applying the Mean Value Theorem, we can proceed to find a point c that satisfies its conclusion.
According to the Mean Value Theorem, 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 point c in (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)
In our case, a = 1, b = 3, and f(x) = x^(-1).
Applying the Mean Value Theorem, we have:
f'(c) = (f(3) - f(1)) / (3 - 1)
To find f(3), substitute x = 3 into the function f(x):
f(3) = 3^(-1) = 1/3
Similarly, to find f(1), substitute x = 1 into the function f(x):
f(1) = 1^(-1) = 1
Now, substituting the values into the equation:
f'(c) = (1/3 - 1) / (3 - 1)
= (-2/3) / 2
= -1/3
To find the value of c, we can solve the equation -1/3 = f'(c):
-1/3 = f'(c)
Differentiate the function f(x) = x^(-1) to find f'(x):
f'(x) = -1 * x^(-2)
= -1 / x^2
Now set f'(c) equal to -1/3 and solve for c:
-1/3 = -1 / c^2
To eliminate the fraction, cross multiply:
- c^2 = 3
Solving for c, take the square root of both sides:
c = Β±β(3)
Therefore, there are two points c that satisfy the conclusion of the Mean Value Theorem for the function f(x) = x^(-1) on the interval [1, 3]. They are c = β(3) and c = -β(3).