Determine if the Mean Value Theorem for Integrals applies to the function f(x) = √x on the interval [0, 4]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.
a) No, the theorem does not apply
b) Yes, x=16/9
c) Yes, x=1/4
d) Yes, x=1/16
x = 16/9
8/3 = 2sqrt(x)
4/3 = sqrt(x)
16/9 = x
To determine if the Mean Value Theorem for Integrals applies to a function, we need to check if the function meets the necessary criteria. The Mean Value Theorem for Integrals states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a value c in (a, b) such that the integral of f(x) from a to b is equal to f(c) times the length of the interval (b-a).
In this case, the function f(x) = √x is a continuous function on the closed interval [0, 4] and differentiable on the open interval (0, 4). Therefore, the Mean Value Theorem for Integrals does apply to this function.
Now, we need to find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem. According to the theorem, there must be a value c in (0, 4) such that the integral of √x from 0 to 4 is equal to √c times the length of the interval (4-0), which is 4.
To find c, we can set up the equation:
∫[0,4] √x dx = √c * 4
We can evaluate the integral of √x from 0 to 4:
∫[0,4] √x dx = [2/3 * x^(3/2)] from 0 to 4
= (2/3 * 4^(3/2)) - (2/3 * 0^(3/2))
= (2/3 * 8) - (2/3 * 0)
= 16/3
Now we can set up the equation:
16/3 = √c * 4
To solve for c, we divide both sides by 4:
(16/3) / 4 = √c
16/12 = √c
Simplifying the fraction on the left side:
4/3 = √c
Squaring both sides, we get:
(4/3)^2 = c
16/9 = c
Therefore, the x-coordinate(s) of the point(s) guaranteed to exist by the Mean Value Theorem for Integrals is x = 16/9.
So, the correct answer choice is b) Yes, x = 16/9.
To determine if the Mean Value Theorem for Integrals applies to the function f(x) = √x on the interval [0, 4], we need to check if the function meets the conditions of the theorem.
The Mean Value Theorem for Integrals 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 number c in the interval (a, b) such that the average value of f on the interval [a, b] is equal to the instantaneous derivative of f at c.
In this case, the function f(x) = √x is continuous on the closed interval [0, 4] and differentiable on the open interval (0, 4), as it is defined and continuous for all values of x in the given interval.
To find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem, we can use the formula for the average value of a function on an interval:
Average value of f on [0, 4] = 1/(4-0) * integral from 0 to 4 of f(x) dx
The integral of √x can be found by using the power rule of integration:
∫√x dx = (2/3)x^(3/2) + C
Plugging in the limits of integration, we have:
Average value of f on [0, 4] = 1/4 * [(2/3)(4)^(3/2) - (2/3)(0)^(3/2)]
Simplifying further, we get:
Average value of f on [0, 4] = (2/3) * (2/3) * 4^(3/2) = (8/9) * 2√2
Now, the instantaneous derivative of f(x) = √x is given by:
f'(x) = 1/(2√x)
To find the x-coordinate(s) that satisfy the Mean Value Theorem for Integrals, we need to find the value(s) of c where f'(c) = (8/9) * 2√2.
Setting the derivative equal to the average value, we have:
1/(2√c) = (8/9) * 2√2
Simplifying, we get:
1/√c = (16/9)√2
Cross-multiplying, we get:
√c = 9/(16√2)
Squaring both sides, we have:
c = (81/256) * 2
Simplifying further, we get:
c = 162/256
Therefore, the correct answer is:
b) Yes, x=16/9
Hence, the Mean Value Theorem for Integrals applies to the function f(x) = √x on the interval [0, 4], and there is at least one point with an x-coordinate of 16/9 guaranteed to exist by the theorem.
what are the conditions that the function must satisfy to apply the MVT?
Does √x meet the conditions on the given interval?
Yes, since √x is continuous on [0,4] and differentiable on (0,4).
(Why is it not differentiable on the closed interval?)
So, the MVT says that there will be some point x=c in (0,4) such that the slope of the line joining the two points on the graph at the ends of the interval is the same as the slope of the tangent line at x=c. That is,
(f(4)-f(0))/(4-0) = f'(c)
That is,
f'(c) = 2/4 = 1/2
So, where is f' = 1/2 ?
f'(x) = 1/(2√x)
so we need 1/(2√c) = 1/2
c=1
Hmmm. Not one of the choices. But, we see from the graphs that
the tangent line at (1,1) does indeed have slope 1/2
https://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9Ax,+y%3D1%2F2+(x-1)%2B1,y%3Dx%2F2+for+0%3C%3Dx%3C%3D4