Given function f defined by f(x) = ( 1- x)³. What are all values of c, in the closed interval [0,3], that satisfy the conditions of the Mean Value Theorem?
The mean value theorem states that a continuous function between x=a and x=b will have at least one tangent parallel to the chord AB.
For f(x)=(1-x)³, the chord between 0 and 3 has a slope of
s=(f(3)-f(0))/(3-0)=-3
The value(s) of c required must satisfy
f'(c)=-3
So, differentiate with respect to x:
f'(x)=-3(1-x)²
and solve for
f'(x)=-3
to get
x=0 and x=2
Reject any solution that is not on the interval [0,3].
mean value theorem of x^3+x-6 with range of [0,2]
To determine the values of c that satisfy the conditions of the Mean Value Theorem (MVT) for the function f(x) = (1 - x)^3 on the closed interval [0,3], we need to check two conditions:
1. Continuity: The function f(x) must be continuous on the closed interval [0,3].
2. Differentiability: The function f(x) must be differentiable on the open interval (0,3).
Let's check each condition step by step.
1. Continuity:
To check continuity, we need to ensure that the function f(x) is continuous on the closed interval [0,3]. A function is continuous if there are no interruptions or sharp corners in its graph.
In this case, the function f(x) = (1 - x)^3 is a polynomial function, and all polynomial functions are continuous for all values of x. Therefore, the function f(x) = (1 - x)^3 is continuous on the interval [0,3].
2. Differentiability:
To check differentiability, we need to ensure that the function f(x) is differentiable on the open interval (0,3). A function is differentiable if it has a derivative at every point in its domain.
To find the derivative of f(x) = (1 - x)^3, we can use the power rule of differentiation:
f'(x) = 3(1 - x)^(3-1) * (-1)
f'(x) = -3(1 - x)^2
The derivative is defined for all values of x, including the open interval (0,3).
Now, we have checked both conditions:
1. The function f(x) = (1 - x)^3 is continuous on the closed interval [0,3].
2. The function f(x) = (1 - x)^3 is differentiable on the open interval (0,3).
Therefore, by the Mean Value Theorem (MVT), there exists at least one value c in the open interval (0,3) such that f'(c) = (f(3) - f(0))/(3 - 0).
Let's compute the values of f(3) and f(0):
f(3) = (1 - 3)^3
f(3) = (-2)^3
f(3) = -8
f(0) = (1 - 0)^3
f(0) = 1^3
f(0) = 1
Now, we can find the value of c by setting:
f'(c) = (f(3) - f(0))/(3 - 0)
-3(1 - c)^2 = (-8 - 1)/3
Simplifying the equation:
-3(1 - c)^2 = -9/3
-3(1 - c)^2 = -3
(1 - c)^2 = 1
1 - c = ±1
Solving for c:
Case 1: 1 - c = 1
-c = 0
c = 0
Case 2: 1 - c = -1
-c = -2
c = 2
Therefore, the values of c, in the closed interval [0,3], that satisfy the conditions of the Mean Value Theorem are c = 0 and c = 2.
In order to find the values of c that satisfy the conditions of the Mean Value Theorem for the function f(x) = (1 - x)³ in the closed interval [0, 3], we need to check if the three conditions of the Mean Value Theorem hold.
The three conditions for the Mean Value Theorem are:
1. Continuity: The function f(x) must be continuous on the closed interval [0, 3].
To check the continuity of f(x) on the interval [0, 3], we need to ensure that the function is defined and there are no holes, jumps, or vertical asymptotes within the interval. In this case, the function f(x) = (1 - x)³ is a polynomial function, and polynomial functions are continuous everywhere. Therefore, f(x) is continuous on the closed interval [0, 3].
2. Differentiability: The function f(x) must be differentiable on the open interval (0, 3).
To check the differentiability of f(x), we need to find its derivative f'(x) and ensure that it is defined and continuous on the open interval (0, 3). Differentiating f(x) = (1 - x)³, we get:
f'(x) = 3(1 - x)² * (-1) = -3(1 - x)²
Since the derivative f'(x) is a polynomial function, it is defined and continuous everywhere. Therefore, f(x) is differentiable on the open interval (0, 3).
3. Equal Average Rate of Change: The function f(x) must have an average rate of change equal to its instantaneous rate of change for at least one value of c in the open interval (0, 3).
To find a value of c that satisfies this condition, we need to find the slope of the secant line between the endpoints of the interval and the function's derivative at that value. The slope of the secant line between two points (x1, f(x1)) and (x2, f(x2)) is given by:
m = (f(x2) - f(x1)) / (x2 - x1)
In our case, x1 = 0 and x2 = 3, so the slope of the secant line is:
m = (f(3) - f(0)) / (3 - 0)
To find f(3), we substitute x = 3 into the function:
f(3) = (1 - 3)³ = (-2)³ = -8
To find f(0), we substitute x = 0 into the function:
f(0) = (1 - 0)³ = 1³ = 1
Now, we can calculate the slope:
m = (-8 - 1) / (3 - 0) = -9/3 = -3
Next, we need to find the derivative of f(x) at some value c in the open interval (0, 3):
f'(x) = -3(1 - x)²
Since we want to find a value of c in the open interval (0, 3), we need to set f'(x) = -3 equal to the slope:
-3 = -3(1 - c)²
Dividing both sides by -3, we get:
1 = (1 - c)²
Taking the square root of both sides:
±√1 = ±(1 - c)
Simplifying:
±1 = ±(1 - c)
Now, we have two cases to consider:
Case 1: 1 = 1 - c
Solving for c, we get:
1 = 1 - c
c = 0
Case 2: -1 = 1 - c
Solving for c, we get:
-1 = 1 - c
-2 = -c
c = 2
Therefore, the values of c that satisfy the conditions of the Mean Value Theorem in the closed interval [0, 3] for the function f(x) = (1 - x)³ are c = 0 and c = 2.