1. Determine whether Rolle's Theorem applied to the function f(x)=((x-6)(x+4))/(x+7)^2 on the closed interval[-4,6]. If Rolle's Theorem can be applied, find all numbers of c in the open interval (-4,6) such that f'(c)=0.
2. Determine whether the Mean Value Theorem applied to the function f(x)=x^2 on the closed interval[4,12]. If Mean Value Theorem can be applied, find all numbers of c in the open interval (4,12) such that f'(c)= f(12)-f(4)/(12-4).
3. Determine whether the Mean Value Theorem applied to the function f(x)=2sinx+sin2x on the closed interval[4pi,5pi]. If Mean Value Theorem can be applied, find all numbers of c in the open interval (4pi,5pi) such that f'(c)= f(5pi)-f(4pi)/(5pi-4pi).
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1. To determine if Rolle's Theorem can be applied to the given function f(x) = ((x-6)(x+4))/(x+7)^2 on the interval [-4, 6], we need to check two conditions:
a) The function must be continuous on the closed interval [-4, 6].
To check continuity, we need to verify that the function is defined and has finite values for every point in the interval. In this case, the function is defined for all values of x except -7 (where the denominator is zero). However, since -7 is not in the interval [-4, 6], the function is continuous on the interval.
b) The function must be differentiable on the open interval (-4, 6).
To check differentiability, we need to ensure that the derivative of the function exists and is finite for every point in the open interval. Let's find the derivative:
f(x) = ((x-6)(x+4))/(x+7)^2
To find the derivative, we can use the quotient rule:
f'(x) = [(x+7)^2(1) - (x-6)(2(x+7))]/(x+7)^4
= [x^2 + 14x + 49 - 2x^2 -2(13)x - 2(42)]/(x+7)^4
= [-x^2 - 14x -49 -26x - 84]/(x+7)^4
= [-3x^2 - 40x - 133]/(x+7)^4
Now, we need to find all values of c in the open interval (-4, 6) such that f'(c) = 0. We set f'(x) equal to zero and solve for x:
-3x^2 - 40x - 133 = 0
You can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Once you find the values of x that make the derivative equal to zero, you will have your values of c. These values of c will satisfy the conditions of Rolle's Theorem.
2. Similar to the previous question, we need to check the conditions for the Mean Value Theorem to be applied to the function f(x) = x^2 on the interval [4, 12].
a) The function must be continuous on the closed interval [4, 12].
The function f(x) = x^2 is a polynomial function, which is continuous for all real numbers. Therefore, it is continuous on the interval [4, 12].
b) The function must be differentiable on the open interval (4, 12).
The function f(x) = x^2 is a polynomial function, and all polynomial functions are differentiable for all real numbers. Therefore, it is differentiable on the open interval (4, 12).
To apply the Mean Value Theorem, we need to find all values of c in the open interval (4, 12) such that f'(c) = (f(12) - f(4))/(12 - 4).
f'(x) = 2x (derivative of x^2)
So, f'(c) = 2c.
Now, we have:
2c = (12^2 - 4^2)/(12 - 4)
2c = (144 - 16)/8
2c = 128/8
c = 64/8
c = 8
Therefore, the value of c that satisfies the Mean Value Theorem is c = 8.
3. In order to determine if the Mean Value Theorem applies to the function f(x) = 2sinx + sin2x on the closed interval [4π, 5π], we need to check the same conditions as before:
a) The function must be continuous on the closed interval [4π, 5π].
The function f(x) = 2sinx + sin2x is a combination of sine functions, which are continuous for all real numbers. Therefore, it is continuous on the interval [4π, 5π].
b) The function must be differentiable on the open interval (4π, 5π).
To check differentiability, we need to find the derivative of the function:
f(x) = 2sinx + sin2x
f'(x) = 2cosx + 2cos2x
The derivative is a combination of cosine functions, which are differentiable for all real numbers. Therefore, the function is differentiable on the open interval (4π, 5π).
Now, we can proceed to find all values of c in the open interval (4π, 5π) such that f'(c) = (f(5π) - f(4π))/(5π - 4π).
f'(c) = 2cosc + 2cos2c
To find the values of c, we need to solve the equation 2cosc + 2cos2c = (2sin5π + sin2(5π) - 2sin4π - sin2(4π))/(5π - 4π).
Simplifying the right side of the equation gives us:
2cosc + 2cos2c = (0 + 0 - 0 - 0)/(π)
= 0/π
= 0
We are solving for the values of c for which the derivative f'(c) equals zero. Since 0 = 0, this equation holds true for all values of c. Therefore, there are infinitely many values of c in the open interval (4π, 5π) such that f'(c) = (f(5π) - f(4π))/(5π - 4π).