If f(x) = ι(x2 - 8)ι, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem? (4 points)


Three


Two


One


None

one

one

To determine how many numbers in the interval satisfy the conclusion of the mean value theorem, we need to check if the function is continuous on the interval and differentiable on the open interval.

First, let's check if the function is continuous on the interval [0, 2.5]. The function f(x) = ι(x^2 - 8)ι is continuous everywhere except where the absolute value function changes sign. The sign of (x^2 - 8) changes at x = -√8 and x = √8. However, these values are not within the interval [0, 2.5], so the function is continuous on the interval.

Next, let's check if the function is differentiable on the open interval (0, 2.5). The function f(x) = ι(x^2 - 8)ι is not differentiable at the points where the absolute value function changes sign, which are x = -√8 and x = √8. Again, these values are not within the open interval (0, 2.5).

Since the function is continuous on the interval and not differentiable at any point within the open interval, there are no numbers in the interval 0 ≤ x ≤ 2.5 that satisfy the conclusion of the mean value theorem. Therefore, the correct answer is "None".

To determine the number of points in the interval 0 ≤ x ≤ 2.5 that satisfy the conclusion of the mean value theorem for the function f(x) = ι(x^2 - 8)ι, we need to first understand the conditions of the mean value theorem.

The mean value theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

In this case, the function f(x) = ι(x^2 - 8)ι is the absolute value of (x^2 - 8).

To find the points that satisfy the mean value theorem, we need to check if f(x) is continuous and differentiable on the interval 0 ≤ x ≤ 2.5.

First, let's check continuity. The absolute value function |x^2 - 8| is continuous everywhere, including the interval from 0 to 2.5.

Next, let's check differentiability. To calculate the derivative of |x^2 - 8|, we can consider the two cases:

1. When x^2 - 8 < 0:
In this case, f(x) = -(x^2 - 8) = -x^2 + 8.
The derivative of -x^2 + 8 is -2x.

2. When x^2 - 8 ≥ 0:
In this case, f(x) = x^2 - 8.
The derivative of x^2 - 8 is 2x.

To determine where the derivative exists, we need to find where both -2x and 2x are defined. Since -2x and 2x are defined for all values of x, the function f(x) = ι(x^2 - 8)ι is differentiable on the interval 0 ≤ x ≤ 2.5.

Now, let's identify the points that satisfy the conclusion of the mean value theorem. It states that there exists at least one number c in the open interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

For this particular question, the interval is 0 ≤ x ≤ 2.5. To find the points that satisfy the conclusion of the mean value theorem, we need to find the number of points in this interval (0, 2.5) where the derivative f'(x) is equal to (f(2.5) - f(0))/(2.5 - 0).

Since f(x) = ι(x^2 - 8)ι, we need to calculate the values of f(x) at the endpoints of the interval (0, 2.5), which are f(0) and f(2.5).

f(0) = ι(0^2 - 8)ι = ι-8ι = 8
f(2.5) = ι(2.5^2 - 8)ι = ι-1.75ι = 1.75

The slope of the secant line connecting f(0) and f(2.5) is (f(2.5) - f(0))/(2.5 - 0) = (1.75 - 8)/2.5 = -6.25/2.5 = -2.5.

Now, we need to find the points in the interval (0, 2.5) where the derivative of f(x) is equal to -2.5.

For f(x) = -x^2 + 8:
-2x = -2.5
Solving for x, we get x = 1.25.

For f(x) = x^2 - 8:
2x = -2.5
Solving for x, we get x = -1.25.

Thus, there is only one number, x = 1.25, in the interval 0 ≤ x ≤ 2.5 that satisfies the conclusion of the mean value theorem for the function f(x) = ι(x^2 - 8)ι.