application of mean value theorem

Question

The function f(x)= x^(2/3)  on [-8,8] does not satisfy the conditions of the mean-value theorem because...

 

A. f(0) is not defined.        

B. f(x) is not continuous on [-8,8].              

C. f ‘ (-1) does not exist.                           

D. f(x) is not defined for x < 0.

E. f ‘ (0)does not exist.

 

Answer 1

This one is a bit confusing, because different teachers have different perspectives.

 

f(x)=(2/3)x^(2/3)

 

f'(x)=(2/3)x^(-1/3)=(2/3)(1/x^1/3)

 

f'(x)<0 when x>0

 

f'(x) is undefined when x=0

 

If your teacher takes the position that the cube root of a negative number is the negative of the cube root of the  corresponding positive number, e.g., cube root of -8=-2=-cube root of 8, then

 

f'(x)>0 when x<0

 

If your teacher takes the position that x^(2/3) for x<0 is a positive number, then f(x)=f(-x), e.g., 8^(2/3)=(8^2)^(1/3)=64^(1/3)=4=(8^(1/3)^2=2^2.

 

In this case, f(-8)=f(8), so the mean value theorem says f'(c)=0 for some -8<c<8.

 

But nowhere is f'(c)=0 for -8<c<8 because f(x) is not continuous.

 

In this case, your teacher should accept choice B as correct.

 

 

 

 

 

Answer 2

f(x) = x^2/3   f'(x) = (2/3)x^-1/3 = 2/(3x^1/3)

 

f'(0) is undefined (division by 0 is not allowed)

 

So, f is not differentiable on (-8, 8).

 

Therefore, the Mean Value Theorem does not apply.  

 

ANSWER:  (E).