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

First let us find the slope of the straight line formed when x1 = 0 to x2 = 2.5.

y = x^2 - 8

y1 = 0^2 - 8 = - 8

y2 = 2.5^2 - 8 = - 1.75

The formula for finding the slope is:

m = (y2 - y1) / (x2 - x1)

m = (-1.75 - ( - 8)) / (2.5 - 0)

m = 2.5

The mean value theorem states that the slope must be 2.5 at least once between x1 = 0 to x2 = 2.5.

Taking the 1st derivative (slope) of the equation:

dy / dx = 2x

Since dy / dx = m = 2.5

2x = 2.5

x = 1.25

Therefore the answer is: One number at x = 1.25