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

1. mathmate Group Title

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 Group Title

the interval is [0,4]

3. oldrin.bataku Group Title

... 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 Group Title

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

5. abccindy Group Title

6. mathmate Group Title

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

7. abccindy Group Title

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

8. mathmate Group Title

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

9. abccindy Group Title

yes to both

10. mathmate Group Title

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

11. mathmate Group Title

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

12. abccindy Group Title

no because it is 0

13. mathmate Group Title

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

14. abccindy Group Title

it is 1-1/ |1-1|

15. mathmate Group Title

Is the derivative really zero?

16. abccindy Group Title

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

17. mathmate Group Title

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

18. mathmate Group Title

...exist and equal.

19. oldrin.bataku Group Title

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 Group Title

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

21. abccindy Group Title

Yes that is correct.

22. mathmate Group Title

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

23. abccindy Group Title

they are all undefined?

24. mathmate Group Title

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.