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

In order to apply the Mean Value Theorem, we first have to check that it is allowed, and there are exactly two conditions. The function has to be continuous on the CLOSED interval [0, 1], and differentiable on the OPEN interval (0, 1). That closed versus open distinction is the whole trick in this problem, so it is worth being careful.

Continuity first. f(x) = x^(2/3) is the cube root of x, squared. Cube roots are defined for every real number, including negatives and zero, so this function is continuous everywhere, and certainly on [0, 1].

Now differentiability:

f'(x) = (2/3)x^(-1/3) = 2/(3 times the cube root of x)

This is undefined at x = 0, because you would be dividing by zero. Here is where students often stop and say the MVT fails. But look carefully at what the theorem asks for. It only requires differentiability on the OPEN interval (0, 1), and x = 0 is not in that open interval. So the one bad point is excluded, and the condition is satisfied.

Both conditions hold, so the MVT applies.

Now we find c. The theorem says there is some c where the instantaneous rate equals the average rate across the whole interval:

f'(c) = (f(1) - f(0))/(1 - 0)

Since f(1) = 1^(2/3) = 1 and f(0) = 0, that right side is just 1/1 = 1. So we need:

(2/3)c^(-1/3) = 1

Rewriting the negative exponent as a fraction:

2/(3 times the cube root of c) = 1

Multiplying both sides by the cube root of c and dividing by 3:

cube root of c = 2/3

Cubing both sides:

c = 8/27

Remember, we should confirm this actually lands inside our interval. 8/27 is about 0.296, which is comfortably between 0 and 1.

So, our final answer is that the Mean Value Theorem does apply, and c = 8/27.

Answer 3

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).