Question:

Verify mean value theorem for the function f(x)=x^3−5x^2−3x, in the interval [a,b], where a=1 and b=3. Find all cϵ(1,3) for which f^1(c)=0.

Answer:

1. First, calculate the derivatives of f(x) as follows: f^1(x) = 3x^2 - 10x - 3

2. Now, set f^1(x) = 0 and solve for the value of x: 3x^2 - 10x - 3 = 0 (3x + 1)(x - 3) = 0 x = -1/3 or x = 3

3. Since the interval of interest is [1,3], the only valid value of x is 3.

4. To verify the mean value theorem, we need to find a value of c such that f^1(c) = 0. Substituting c = 3 into f^1(x) gives us: f^1(3) = 33^2 - 103 - 3 = 0

5. Therefore, c = 3 is a valid value of c that satisfies the mean value theorem for the given function and interval.

Question:

Examine the applicability of Mean Value Theorem for the following function. f(x)=x^2−1 for xϵ[1,2]

Answer:

1. Mean Value Theorem states that if a function is continuous on a closed interval [a, b], then there exists a number c between a and b, such that the function’s derivative at c is equal to the average rate of change of the function over the interval.

2. Since the function f(x)=x^2−1 is continuous on the interval [1,2], it satisfies the condition of Mean Value Theorem.

3. Therefore, Mean Value Theorem is applicable to the given function.

Question:

Verify Mean Value Theorem, if f(x)=x^2−4x−3 in the interval [a,b], where a=1 and b=4.

Answer:

Mean Value Theorem states that if f(x) is continuous on the interval [a,b] and differentiable on (a,b), then there exists a c in (a,b) such that f’(c) = (f(b) - f(a)) / (b-a).

Step 1: Find f’(x) f’(x) = 2x - 4

Step 2: Calculate f’(c) f’(c) = 2c - 4

Step 3: Calculate f(a) and f(b) f(a) = 1^2 - 4(1) - 3 = -4 f(b) = 4^2 - 4(4) - 3 = -3

Step 4: Calculate (f(b) - f(a)) / (b-a) (f(b) - f(a)) / (b-a) = (-3 - (-4)) / (4 - 1) = -7/3

Step 5: Set f’(c) = (f(b) - f(a)) / (b-a) 2c - 4 = -7/3

Step 6: Solve for c 2c = -7/3 + 4 2c = 7/3 c = 7/6 = 1.167

Question:

What went wrong when the tortoises, snakes and lizards left the forest?

Answer:

1. The tortoises, snakes and lizards left the forest.
2. It is unclear what went wrong after they left.

Question:

Examine if Rolles theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolles theorem from these example ? (i) f(x)=[x] for x∈[5,9] (ii) f(x)=[x] for x∈[−2,2] (iii) f(x)=x^2−1 for x∈[1,2]

Answer:

(i) Yes, Rolles theorem is applicable to the function f(x)=[x] for x∈[5,9]. The converse of Rolles theorem states that if a function is continuous at a point and its derivative is zero at that point, then the function has a local maximum or minimum at that point. Thus, for this function, there may be a local maximum or minimum at the point x=5 or x=9.

(ii) Yes, Rolles theorem is applicable to the function f(x)=[x] for x∈[−2,2]. The converse of Rolles theorem states that if a function is continuous at a point and its derivative is zero at that point, then the function has a local maximum or minimum at that point. Thus, for this function, there may be a local maximum or minimum at the point x=-2 or x=2.

(iii) No, Rolles theorem is not applicable to the function f(x)=x^2−1 for x∈[1,2] because the function is not continuous at the point x=1 and x=2. Therefore, the converse of Rolles theorem does not apply.

Question:

Verify Rolles Theorem for the function f(x)=x^2+2x−8,x∈[−4,2].

Answer:

Step 1: First, we need to check whether the function f(x) is continuous in the given interval [−4,2].

Step 2: Then, we need to check whether the function f(x) has a local maximum or minimum value in the given interval.

Step 3: After that, we need to calculate the value of f(x) at the endpoints of the interval, i.e., f(-4) and f(2).

Step 4: Finally, we need to compare the value of f(x) at the endpoints with the local maximum or minimum value of f(x). If the value of f(x) at the endpoints is equal to the local maximum or minimum value of f(x), then Rolles Theorem is verified.

Question:

Roll’s theorem is applicable on following function. If this statement is true enter 1 otherwise 0. f(x)=x^2 −1 for x ϵ [1,2]

Answer:

Answer: 1

Question:

Examine the Rolles theorem is applicable to the followng function. Find the number of points the following function is not continous? f(x)=[x] for x ϵ [2,2]

Answer:

Step 1: Rolles theorem states that if a function f(x) is continuous on the interval [a,b] and differentiable on the interval (a,b) and if f(a)=f(b) then there exists at least one c in (a,b) such that f’(c)=0.

Step 2: In the given problem, the function f(x)=[x] is defined on the interval [2,2]. Since the function is not differentiable on the interval (2,2), Rolles theorem is not applicable.

Step 3: Therefore, the number of points at which the function is not continuous is 0.

Question:

Examine the applicability of Mean Value Theorem for the following function. f(x)=[x] for x ϵ [2,2]

Answer:

1. The Mean Value Theorem states that if a function f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one c in the interval (a,b) such that f’(c) = (f(b) - f(a))/(b-a).

2. To apply the Mean Value Theorem to the given function, we must first determine if the function is continuous and differentiable on the open interval (2,2).

3. Since the function is not defined on the open interval (2,2), the Mean Value Theorem cannot be applied.