## finaynay 3 years ago Find the point(s) guaranteed by the mean value theorem for f(x)=|x^2-9| over the interval [0,2]

1. math>philosophy

x^2 - 0?

2. finaynay

i meant 9 lol

3. math>philosophy

lol

4. math>philosophy

So what are the conditions for the MVT

5. finaynay

well it is already guaranteed that the mean value theorem applies so i did f'(x)= f(2)-f(0)/2 f'(x)=-2 then what do i find the derivative of f(x)=|x^2-9| because i dont know how to find the derivative of a absolute value

6. math>philosophy

$\sqrt{x^2} = |x|$

7. finaynay

how would i do that for |x^2-9| and did i do the beginning of the problem correctly

8. Callisto

y =|x^2-9| Let u =x^2-9 => $$y=\sqrt{ u^2}$$ $\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$ Does that help?

9. finaynay

i still don't understand

10. Callisto

Which part?

11. finaynay

how i'm supposed to take the derivative of $|x ^{2}-9|$

12. Callisto

Do you know that $$\sqrt{a^2} = |a|$$?

13. finaynay

yes but how do i apply that $\sqrt{(x-9)^{2}}$ isn't the derivative

14. Callisto

It's the question :P To find the derivative of |x^2 -9|, you can find the derivative of $$\sqrt{(x^2-9)^2}$$, which is the same as |x^2 -9|. Btw, it should be $$\sqrt{(x^2-9)^2}$$, not $$\sqrt{(x-9)^2}$$

15. finaynay

oh okkkkkkk but is that even what i'm supposed to be doing in this problem

16. Callisto

You need to find f'(x), so, I think so :|

17. finaynay

for realsssss noooooooo

18. Callisto

Hmmm.. MVT: Suppose y=f(x) is continuous on a closed interval [a,b] and differentiable on the interior (a,b). Then there is at least one point c at which $$\frac{f(b)-f(a)}{b-a} = f'(c)$$ In this question, the conditions are satisfied, that is f(x) = |x^2-9| is continuous on the closed interval [0, 2] and is differentiable on (0, 2) So, just apply it: $$\frac{f(2)-f(0)}{2} = f'(c)$$ And you have to find f'(x), isn't it?

19. finaynay

yeah

20. Callisto

So... just differentiate that guy :|

21. finaynay

so the derivative is 2x

22. finaynay

x=-1?

23. satellite73

not quite

24. satellite73

on the interval $$[0,2]$$ you have $$|x^2-9|=9-x^2$$

25. finaynay

how is it 9−x^2

26. satellite73

since this is a quadratic, the "point guaranteed to exist by the mean values theorem will be in the midpoint of the interval, namely at $$x=1$$

27. satellite73

if $$-3<x<3$$ we have $$x^2-9<0$$ and so $$|x^2-9|=-(x^2-9)=9-x^2$$

28. satellite73

in english $$x^2-9$$ is negative on the interval $$[0,2]$$ and so the absolute value is minus the whole thing , namely $$9-x^2$$

29. finaynay

okay so the derivative is -2x right and then you set that equal to the mean and you solve for x and you get x=1?

30. satellite73

lets try it

31. satellite73

$$f(0)=9$$ and $$f(2)=5$$ and so $$\frac{f(2)-f(0)}{2-0}=\frac{5-9}{2}=-4$$

32. finaynay

isn't it -2

33. satellite73

34. finaynay

oh okay lol

35. math>philosophy

<3

36. satellite73

yes it is $$-2$$ now we set $$-2x=-2$$ and guess what? $$x=1$$

37. finaynay

-2=f'(x)

38. finaynay

yessss! alright i get it now

39. satellite73

you can convince yourself, if you have the time in inclination, that the "point guaranteed to exist by the mean value theorem" is smack in the middle of the interval if your function is a quadratic