Determine the value for c on {2, 5} that satisfies the mean value theorem for f(x) = x^2 - 3 / x -1

a. 2
b. 3
c. -1

show all work

B. 3.

Step-by-step explanation:

OK lets try again.

The slope of the secant = slope of the tangent at a certain point ( The Mean Value Theorem).

Slope of the secant = f(5) - f(2) / (5 - 2)

= [(25-3) / (5-1)  - (4-3) / (2-1)] / 3

=  (22/4 - 1) / 3

= 9/2 / 3

= 9/6

= 3/2.

The derivative at c = the slope of the tangent at c.

Finding the derivative:

f'(x) =  [2x(x - 1) - (x^2 - 3) ]/ (x - 1)^2    (where x = c).

= (x^2 - 2x + 3)/ (x - 1)^2 = the slope.

So equating the slopes:

(x^2 - 2x + 3) / (x - 1)^2 = 3/2

2x^2 - 4x + 6 = 3x^2 - 6x + 3

x^2 - 2x - 3 = 0

(x - 3)(x + 1) = 90

x = 3 , -1

x can't be -1 because we are working between the values 2 and 5 so

x = c = 3.