mean value theorem

Question

let f(x)= x³ + 3x² -10x , so that f(1)= -6 and f(2)= 0. by the mean value theorem, there exist a number y in the open interval (0,1) such that f ’(1+y) = 6. the value of y turns out to be 

a) greater than 3/4

b) between 1/2 and 3/4

c) between 1/4 and 1/2

d) less than 1/4

 please help me explain the answer, plzzzzzz

Answer 1

f(x) = x³ + 3x² - 10x

 

f'(x) = 3x² + 6x - 10

 

f(x) is continuous on [1, 2] and is differentiable on (1, 2).  So, by the Mean Value Theorem, there is at least one number, c, in the interval   (1, 2) so that f'(c) = [f(2) - f(1)]/(2-1) = [0 - (-6)]/ 1 = 6.

 

So, 3c² + 6c - 10 = 6

 

     3c² + 6c - 16 = 0

 

                     c = [-6 ± √(36 + 192)]/6 = [-6 ± 15.1]/6 ≈ 1.52, -1.65

 

f'(c) = 6 for c ≈ 1.52 and 1 < c < 2

 

Let c = 1+y   Then, 1 + y = 1.52   

 

                                  y = .52 which is between 1/2 and 3/4

 

f'(1+y) = f'(c) = 6