## abccindy Group Title does the mean value theorem apply to |x-1|? one year ago one year ago

1. mathmate

The mean value theorem applies to a specified closed interval, and it must be differentiable within the (open) interval. This should help you put restriction on whether the MVT applies to |x-1| or not.

2. abccindy

the interval is [0,4]

3. oldrin.bataku

... you haven't given us enough information. It must be continuous over $$[a,b]$$ and differentiable over $$(a,b)$$, i.e. $$0\notin(a,b)$$.

4. oldrin.bataku

Ok, you tell us: is $$|x-1|$$ continuous on $$[a,b]$$ and differentiable on $$(a,b)$$?

5. abccindy

6. mathmate

Hint: |dw:1358042258899:dw| and try to answer questions by oldrin.

7. abccindy

i got to that part but does the vertical asymptote have anything to do with it?

8. mathmate

The main questions are: is f(x) continuous on (0,4), and is f(x) differentiable on [0,4]

9. abccindy

yes to both

10. mathmate

@geerky42 are we able to determine the derivative of f(x)=|x-1| at x=1?

11. mathmate

@abccindy same question for you then: are we able to determine the derivative of f(x)=|x-1| at x=1?

12. abccindy

no because it is 0

13. mathmate

If it is zero, then the answer is yes, because zero is a number.

14. abccindy

it is 1-1/ |1-1|

15. mathmate

Is the derivative really zero?

16. abccindy

$f'(x)=\frac{ x-1 }{ |x-1| }$

17. mathmate

For the derivative to exist, the left and right limits of f(x) must be equal.

18. mathmate

...exist and equal.

19. oldrin.bataku

The derivative is undefined... c'mon people! There are two different limits of for the slope at x=1 depending on whether you approach from the left or right.

20. mathmate

Would you propose the left limit, i.e. f'(1-), and the right limt f'(1+).

21. abccindy

Yes that is correct.

22. mathmate

So if the derivative is undefined at x=1, what can we conclude if the MVT applies to the interval (0,4)?

23. abccindy

they are all undefined?

24. mathmate

Yes, the derivative at x=1 is undefined because f'(1-)=-1, f'(1+)=+1, so f'(1-)$$\ne$$f'(1+), by definition of the derivative, it is undefined at x=1. This concept is even more important when you work with multi-variable calculus in the near future.