Verify the hypothesis of the mean value theorem for each function below defined on the indicated interval. Then find the value “C” referred to by the theorem.
Q1a) h(x)=√(x+1 ) [3,8]
Q1b) K(x)=(x-1)/(x=1) [0,4]
Q1c) Explain the difference between the Mean Value Theorem and Rollo’s Theorem.
Rolle's theorem is just the mean value theorem, where f(x) = 0 at both endpoints.
What did you get to parts a and b?
I get stuck at this and don't know how to from here.... can you help...
Q1a) h(x)=√(x+1 ) [3,8]
MVT=[h(b)-h(a)]/(b-a)=h'(c)
To find h(b) and h(a), we just plug endpoints into original function
h(b)=h(8)=√(x+1 )
h(b)=h(8)=√(8+1 ) = 3
h(a)=h(3)= √(3+1 ) = 2
MVT=[3-2]/[8-3] =f^' (c)
MVT=1/2=f^' (c)
Next, we find the derivative for h(x)
h'(x)=√(x+1 )
h'(x)=(d/dx(x+1))/(2√(x+1))
h'(x)=(1+0)/(2√(x+1))
h'(x)=1/(2√(x+1))
h'(c)=h'(x)
1/2=(1+0)/(2√(x+1))
if h(x) = √(x+1)
h'(x) = 1/(2√(x+1))
so, we want c where
h'(c) = (3-2)/(8-3) = 1/5
1/2√(x+1) = 1/5
5 = 2√(x+1)
√(x+1) = 5/2
x+1 = 25/4
x = 21/4
and 3 < 21/4 < 8
-----------------------------
if k(x) = (x-1)/(x+1)
k'(x) = 2/(x+1)^2
k(0) = 0
k(4) = 3/5
so, we want c where k'(c) = (3/5)/4 = 3/20
3/20 = 2/(x+1)^2
3(x+1)^2 = 40
(x+1)^2 = 40/3
x = -1 + 2√(10/3) = 2.65
0 < 2.65 < 4, so we're ok.
actually, I think Rolle's Theorem is the MVT where f(a) = f(b), so that f'(c) = 0.
oops. k(0) = -1
adjust the calculation accordingly.
To verify the hypothesis of the Mean Value Theorem (MVT) for a function defined on a given interval and find the value "C" referred to by the theorem, we need to follow these steps:
Step 1: Check if the function satisfies the conditions of the Mean Value Theorem. The MVT states that for a function f(x) defined on [a, b], the following conditions must be met:
- f(x) must be continuous on the closed interval [a, b],
- f(x) must be differentiable on the open interval (a, b).
If a function satisfies these conditions, we can proceed to the next step.
Step 2: Find the average rate of change of the function between the endpoints of the interval. The average rate of change is calculated by the formula:
Average rate = (f(b) - f(a)) / (b - a)
Step 3: Find the derivative of the function f'(x).
Step 4: Equate the average rate of change to the derivative of the function:
Average rate = f'(C), where C is a value between a and b.
Step 5: Solve the equation to find the value of C.
Let's apply these steps to the given functions:
Q1a) h(x) = √(x + 1) on the interval [3, 8]
- Step 1: The function h(x) = √(x + 1) is continuous and differentiable on the given interval.
- Step 2: Average rate of change = (h(8) - h(3)) / (8 - 3) = (√9 - √4) / 5 = (3 - 2) / 5 = 1/5.
- Step 3: h'(x) = 1 / (2√(x + 1)) = 1 / (2√(x + 1)).
- Step 4: Set average rate = f'(C). We have 1/5 = 1 / (2√(C + 1)).
- Step 5: Solve for C: 2√(C + 1) = 5. Squaring both sides gives 4(C + 1) = 25, which simplifies to 4C + 4 = 25. Solving for C gives C = 21/4.
Therefore, according to the Mean Value Theorem, there exists a value c in the interval [3, 8] such that h'(c) = 1/5. In this case, c = 21/4.
Q1b) K(x) = (x - 1) / (x = 1) on the interval [0, 4]
- Step 1: The function K(x) = (x - 1) / (x = 1) is not continuous at x = 1, so it does not satisfy the conditions of the Mean Value Theorem. Therefore, the Mean Value Theorem cannot be applied.
Q1c) The Mean Value Theorem (MVT) states that for a function f(x) defined on a closed interval [a, b], if the function satisfies the conditions of being continuous on [a, b] and differentiable on (a, b), there exists a value c in the interval (a, b) such that the derivative of the function f'(c) is equal to the average rate of change of the function between the endpoints of the interval.
Rolle's Theorem, on the other hand, is a special case of the Mean Value Theorem. Rolle's Theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), and if f(a) = f(b), then there exists at least one value c in (a, b) where the derivative f'(c) is equal to zero. In other words, Rolle's Theorem guarantees the existence of at least one point within the interval where the instantaneous rate of change (the derivative) is zero.