Determine if the Mean Value Theorem for Integrals applies to the function f of x equals the square root of x on the interval [0, 4]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem.

x = 16/9

8/3 = 2sqrt(x)
4/3 = sqrt(x)
16/9 = x

it's not 4/3 but you're close Steve, you just needed to do one more step

Well, let's see if the Mean Value Theorem for Integrals applies to the function f(x) = √x on the interval [0, 4].

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 point c in (a, b) such that the average value of f(x) on [a, b] is equal to the instantaneous rate of change of f(x) at c.

In this case, the function f(x) = √x is continuous on the closed interval [0, 4], and it is differentiable on the open interval (0, 4). So the Mean Value Theorem for Integrals applies!

To find the x-coordinate(s) guaranteed to exist by the theorem, we can use the formula:

c = (1/(b-a)) * ∫[a, b] f(x) dx

Applying this formula to our function f(x) = √x over the interval [0, 4], we have:

c = (1/(4-0)) * ∫[0, 4] √x dx

To integrate √x, we get:

c = (1/4) * (2/3)x^(3/2) | from 0 to 4

Simplifying, we have:

c = (1/4) * [(2/3)(4)^(3/2) - (2/3)(0)^(3/2)]

c = (1/4) * [(2/3)(8) - (2/3)(0)]

c = (1/4) * [(16/3) - 0]

c = (1/4) * (16/3)

c = 4/3

So, by the Mean Value Theorem for Integrals, there exists at least one point c in the interval (0, 4) such that f'(c) = 4/3. The x-coordinate of this point is 4/3.

Therefore, the x-coordinate(s) guaranteed to exist by the theorem is/are 4/3.

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 two conditions:

1. Continuity: The function must be continuous on the interval [0, 4]. In this case, f(x) = √x is continuous for all x ≥ 0, so the continuity condition is satisfied since the interval [0, 4] is within the domain of f(x).

2. Integrability: The function must be integrable on the interval [0, 4]. In other words, the function must be defined and bounded on [0, 4]. Since f(x) = √x is defined and bounded on [0, 4], the integrability condition is also satisfied.

Therefore, the Mean Value Theorem for Integrals applies to the function f(x) = √x on the interval [0, 4].

Now, let's find the x-coordinate(s) of the point(s) guaranteed to exist by the theorem. According to the Mean Value Theorem for Integrals, there exists at least one value c in the interval [0, 4] such that:

∫[0, 4] f(x) dx = f(c) * (b - a),

where [a, b] is the interval [0, 4] and f(c) is the value of the function at c.

In this case, we have:

∫[0, 4] √x dx = f(c) * (4 - 0).

To find the x-coordinate(s) of c, we need to evaluate the integral on the left side of the equation:

∫[0, 4] √x dx = ∫[0, 4] x^(1/2) dx.

Using the power rule of integration, we can integrate this function as follows:

= [2/3 * x^(3/2)] evaluated from 0 to 4
= 2/3 * (4^(3/2)) - 2/3 * (0^(3/2))
= 2/3 * 8 - 2/3 * 0
= 16/3.

Now, let's solve for f(c):

16/3 = √c * (4 - 0).
16/3 = √c * 4.
4/3 = √c.

To isolate c, we square both sides of the equation:

(4/3)^2 = (√c)^2.
16/9 = c.

Therefore, the x-coordinate(s) of the point(s) guaranteed to exist by the Mean Value Theorem for Integrals is c = 16/9.

In summary, the Mean Value Theorem for Integrals applies to f(x) = √x on the interval [0, 4], and the x-coordinate(s) of the point(s) guaranteed to exist by the theorem is c = 16/9.

check that f(x) satisfies the theorem's conditions:

yes, f(x) is continuous on [0,4] so we can go on.

the MVT states that there is some c in [0,4] such that
f(c) = 1/(4-0)*∫[0,4] f(x) dx

I get c = 4/3, which is indeed in [0,4]