# Mean Value Theorem

1. Apr 2, 2010

### npellegrino

1. Use the Mean Value Theorem
f(x) = sqrtX - 2x at [0,4] so a = 0, b = 4

3. So I found the derivative (which is the slope) and then set the derivative equal to the 1 because of f(b)-f(a)/b-a
f'(x) = 1/2sqrtX - 2 so 1/2sqrtX - 2 = 0, then 1/2sqrtX = 2, then my solution is x = 1/16

Just would like a heads up if this is correct, thanks in advanced.

2. Apr 2, 2010

### Staff: Mentor

No, this isn't correct. I think you probably made a mistake in your calculation of (f(b) - f(a))/(b - a), so check your work on this calculation.

Notice also that the extra parentheses I used (and you didn't) are necessary. The correct interpretation of what you wrote would be
$$f(b) - \frac{f(a)}{b} - a$$

and I'm sure that's not what you intended.

3. Apr 2, 2010

### npellegrino

well the Mean Value Theorem is

so if a = 0 and b = 4 wouldn't the solution be 1 ? then you set the derivative of the function to 1 which gave me 1/16 ?

4. Apr 2, 2010

### Staff: Mentor

I think you might be calculating f'(a) and f'(b), instead of f(a) and f(b). Anyway (f(b) - f(a))/(b - a) $\neq$ 1.

5. Apr 2, 2010

### cheddacheeze

subsitute the values of a and b into the formula
find the derivative of f(x)
and after youve got f'(x) and (f(b)-f(a))/b-a

6. Apr 2, 2010

### npellegrino

the derivative i got for f(x) is 1/2sqrtX - 2 then i set that equals to (f(b)-f(a))/b-a which is 1, so i'm trying to figure 1/2sqrtX - 2 = 1 and i'm having trouble figuring the fraction

7. Apr 2, 2010

### cheddacheeze

how are you getting the slope = 1, i got -6/4
f(4) = 4^1/2 - 8 = -6
f(0) = 0
(-6-0)/(4-0) does not equal to 1

well using my values i get 1/(2(x^1/2)) = -6/4+2
you can probably work your way from there

8. Apr 2, 2010

### Staff: Mentor

Stop already! (f(b)-f(a))/(b-a) IS NOT EQUAL TO 1 !

Put in the actual numbers for a and b and do the calculation!

9. Apr 2, 2010

### npellegrino

You are not taking the derivatives of f(b) and f(a)... the mean value theorem states that if f is continuous on the closed intervals [a,b] which in this case is [0,4] and differentiable on the open interval (a,b,) then there exists a number c in (a,b) such that

10. Apr 2, 2010

### npellegrino

maybe my professor has taught me wrong?

11. Apr 2, 2010

### cheddacheeze

just look at the formula and substitute a and b... into the equation

12. Apr 2, 2010

### npellegrino

Alright i'm having a brainfart i see what you are saying.

13. Apr 2, 2010

### npellegrino

f(4) = sqrt4 - 8 = -6
f(0) = sqrt0 - 2(0) = 0

14. Apr 2, 2010

### npellegrino

Alright i think i solved it... finally thanks for the help i really do appreciate it from you 2 :)

f(4) = -6
f(0) = 0
which gives me -6/4

the derivative 1/2sqrtX -2 = -6/4 which will give me 1/2sqrtx = 1/2 then x = 1

15. Apr 2, 2010

### Staff: Mentor

Yes.

16. Apr 2, 2010

### Staff: Mentor

I'm pretty sure he didn't teach you that (2 - 8)/(4 - 0) = 1.

17. Apr 2, 2010

### cheddacheeze

not so sure about that, sometimes the things they teach you are so obstract that they themselves get confused

18. Apr 2, 2010

### Staff: Mentor

No question that an instructor can occasionally get confused with some abstract calculation, but (2 - 8)/(4 - 0) in no way can be considered as abstract.

19. Apr 2, 2010

### cheddacheeze

indeed it is not abstract, the mean value theorem is very elementary since its dealing with slopes and derivatives so its easy to understand