Mean value theorem

Question

Determine whether the Mean Value Thereom can be applied to f on the closed interval (a,b) .  If the MVT can be applied, find all values of c in the open interval (a,b) such that f'(c) =f(b)-f(a) / b - a.

f(x) = x 2/3, (0,1)

Note it's x to the 2/3 power

Answer 1

Step 1 Differentiate the given function                   f(x)=x^2/3

         Use the power rule.                                   f'(x)=2/3x^-1/3

Step 2 Evaluate the derivative of c.                      f'(c)=2/3c^-1/3

Step 3 Find the slope over the entire interval.      

          Evaluate the function at b=1.                    f(b)= f(1)= 1^2/3= 1

          Evaluate the function at a=0.                    f(a) = f(0)= 0^2/3 = 0

         Substitute in slope formula.                       f(b)-f(a) / b-a =1-0 / 1-0

Step 4 Set the derivative (Step 2) equal to the slope (step 3).     2/3c^-1/3= 1

                                                                                              c^-1/3= 3/2

          Raise both sides to the power -3.                                    (c^-1/3)^-3=(3/2)^-3

          Simplify.                                                                       c= (2/3)³

                                                                                              c= 8/27   

 

                       Hope this helps.

 

Answer 2

If the interval is closed the notation should be [0,1].  In that case, the MVT would not apply because f'(0) is not defined. 

If the interval is closed, and includes a value for 0<a<1, then we can find an appropriate c to satisfy MVT.

Let's try the interval [1/8, 1]

For f(x) = x^2/3, f(1/8) = 1/4 and f(1) = 1.

(f(1) - f(1/8))(1- 1/8)  =  (1 - 1/4)/(1 - 1/8)  = (3/4)/(7/8)  =  6/7

f'(x) = (2/3) x^-1/3  =  2/(3x^1/3)

We need to find c such that f'(c)  =  6/7.

     2/(3c^1/3) = 6/7

      2  =  (6/7) 3c^1/3 = (18/7) c^1/3

       c^1/3  =(2//18/7) = 7/9

 

      ( c^1/3 )³  =  (7/9)³

          c  =  343/729

The value of the derivative at x = c = (7/9)³ is 6/7, which is the slope of the line between f(a) and f(b).