zeesbrat3 one year ago If f(x) = ∣(x2 − 8)∣, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem?

1. zeesbrat3

2. amoodarya

if u plot it , it will be simpler to do

3. amoodarya

$|x^2-8|=|(x-2\sqrt{2})(x+2\sqrt{2})|$

4. amoodarya

|dw:1436664302388:dw|

5. amoodarya

|dw:1436664349288:dw|

6. amoodarya

|dw:1436664378196:dw|

7. zeesbrat3

So, 2 because those are the 0s?

8. zeesbrat3

@amoodarya

9. amoodarya

can you tell me what kind of point , theorem want ?

10. zeesbrat3

You're looking for a point that is defined and continuous on the supplied interval as well as differentiable... right?

11. amoodarya

yes

12. misty1212

HI!! (and holy moly!!0

13. misty1212

you are not looking for a point that is defined and continuous, that doesn't even make any sense !!

14. misty1212

first you have to check to see if your function is differentiable on the interval you are given

15. amoodarya

|dw:1436664956891:dw| does it make sense ?

16. misty1212

since $|x^2-8|=-x^2+8$ on your interval, it is definitely continuous and differentiable (because it is a polynomial)

17. misty1212

you are not asked do find any number here, just how many numbers will satisfy the conclusion of the mean value theorem

18. misty1212

since your function is a quadratic, the derivative is a line when you solve a linear equation there is only one solution

19. misty1212

in fact you can check that the solution to the number guarantee to exist by the mean value theorem is in the center of the interval that is always the case with a quadratic

20. zeesbrat3

21. misty1212

you are welcome, i hope it is clear (more or less) and to reiterate (not to be mean) "You're looking for a point that is defined and continuous on the supplied interval as well as differentiable... right? makes zero sense a "point" is a number a number is always defined, it is a number a number is not "continuous" that is a local property of a function, not a number nor is a number "differentiable" a function may or may not be differentiable on an interval, i.e. the derivative may or may not exist for all numbers in the interval

22. ybarrap

If a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that $$f'(c) = \frac{f(b) - f(a)}{b-a}$$ |dw:1436665525899:dw| https://en.wikipedia.org/wiki/Mean_value_theorem https://www.wolframalpha.com/input/?i=plot+-x%5E2+%2B+8%2C0%3C%3Dx%3C%3D2.5