Determine if the Mean Value Theorem for Integrals applies to the function f(x)=2-x^2 on the interval [0,√2). If so, find the x-coordinates of the point(s) guaranteed by the theorem
a) No, the Mean Value Theorem for Integrals does not apply
b) Yes, x=4/3
c) Yes, x= √5/3
d) Yes, x=√(2/3)
integral = 2 x - x^3/3 at sqrt 2 - at 0
= 2 * 2^.5 - 2^(1.5)/3 - 0
length of rectangle = 2^.5
so
y * [ 2^.5 ] = 2 * 2^.5 - 2^(1.5)/3
y * [ 2^.5 ] = 2 * 2^.5 - 2^.5 *2 /3
y * [ 2^.5 ] = 2 * 2^.5 [ 1-1/3)
y * [ 2^.5 ] =(4/3) (2^.5)
y = 4/3
where is that?
2-x^2 = 4/3
x^2 = 2/3
d
Well, well, well, let's see if the Mean Value Theorem for Integrals applies to our friend f(x) = 2 - x^2 on the interval [0, √2).
According to this theorem, if a function is continuous on a closed interval [a, b], then there exists at least one number c in the interval (a, b) such that the average value of the function over that interval is equal to the value of the function at c.
Now, let's check if our function is continuous on [0, √2). Hmm... looks like it's smooth sailing, no nonsensical jumps or holes to be found. So, yes, our function is continuous here.
Great! So now, we need to find the average value of the function on the interval [0, √2) and then determine if there exists a point c in this interval where the value of the function equals the average value.
To find the average value of f(x), we need to calculate the definite integral of f(x) over the interval [0, √2) and then divide it by the length of the interval (√2 - 0).
∫[0, √2) (2 - x^2) dx
Solve this integral and divide it by (√2 - 0), and you'll get the average value.
But wait, there's more! We need a point c where f(c) is equal to this average value.
Now, I must apologize for my previous clownish antics. I got caught up in my foolery and failed to realize that the Mean Value Theorem for Integrals does not actually provide us with an exact x-value. Therefore, none of the options provided (b, c, or d) are correct.
So, after all that, the correct answer is a) No, the Mean Value Theorem for Integrals does not apply.
I hope my ramblings at least provided you with some entertainment, even if the Mean Value Theorem did not cooperate.
To determine if the Mean Value Theorem for Integrals applies to the function f(x)=2-x^2 on the interval [0,√2), we need to check if the function satisfies the necessary conditions of the theorem.
The Mean Value Theorem for Integrals states that if a function f is continuous on the interval [a,b] and differentiable on the interval (a,b), then there exists at least one value c in the interval (a,b) such that the integral of f(x) over the interval [a,b] is equal to f(c) times the length of the interval (b-a).
In this case, the function f(x)=2-x^2 is continuous on the interval [0,√2) since it is a polynomial. We also need to check if the function is differentiable on the interval (0,√2).
Taking the derivative of f(x), we get f'(x)=-2x. This derivative is defined and continuous on the interval (0,√2), which means f(x) is differentiable on this interval.
Therefore, the function f(x)=2-x^2 satisfies the necessary conditions of the Mean Value Theorem for Integrals. This means the theorem applies to this function on the interval [0,√2).
To find the x-coordinate(s) of the point(s) guaranteed by the theorem, we need to find the value of c that satisfies the equation:
∫[0,√2) (2-x^2) dx = (2-x^2)|[0,√2) = f(c)(√2-0)
To evaluate the integral, we integrate f(x) with respect to x:
∫[0,√2) (2-x^2) dx = [2x-(x^3)/3] |[0,√2) = (2(√2)-((√2)^3)/3) - 0 = 2√2 - (√2)^3/3 = 2√2 - 2/3
So we have:
2√2 - 2/3 = f(c)(√2-0)
Simplifying further:
f(c) = (2√2 - 2/3)/(√2)
f(c) = (2√2/√2) - (2/3√2)
f(c) = 2 - (2/3√2)
To find the value of c, we need to solve the equation f(c) = 2 - (2/3√2) for c.
Since f(c) is a decreasing function, we can look for the solution by setting f(c) equal to the lower endpoint of the interval, which is 0.
0 = 2 - (2/3√2)
2/3√2 = 2
√2/√2 = 3
√2/3 = 1
Therefore, the x-coordinate of the point guaranteed by the Mean Value Theorem for Integrals is c=√2/3.
Therefore, the correct answer is d) Yes, x=√(2/3).
To determine if the Mean Value Theorem for Integrals applies to the function f(x)=2-x^2 on the interval [0,√2), we need to check if the function satisfies the three conditions required for the theorem:
1. Continuity: The function f(x)=2-x^2 is continuous on the interval [0,√2) because it is a polynomial function, and polynomial functions are continuous everywhere.
2. Differentiability: The function f(x)=2-x^2 is differentiable on the interval (0,√2) because it is a polynomial function. However, it is not differentiable at x=0 because the derivative is not defined at that point. Since the theorem requires differentiability on the closed interval [0,√2], the function does not satisfy this condition.
3. Integrability: The function f(x)=2-x^2 is integrable on the interval [0,√2) because it is continuous on this interval.
Since the function does not satisfy the condition of differentiability on the closed interval, the Mean Value Theorem for Integrals does not apply to the function f(x)=2-x^2 on the interval [0,√2).
Therefore, the correct answer is option a) No, the Mean Value Theorem for Integrals does not apply.