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

1. math>philosophy Group Title

x^2 - 0?

2. finaynay Group Title

i meant 9 lol

3. math>philosophy Group Title

lol

4. math>philosophy Group Title

So what are the conditions for the MVT

5. finaynay Group Title

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 Group Title

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

7. finaynay Group Title

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

8. Callisto Group Title

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 Group Title

i still don't understand

10. Callisto Group Title

Which part?

11. finaynay Group Title

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

12. Callisto Group Title

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

13. finaynay Group Title

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

14. Callisto Group Title

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 Group Title

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

16. Callisto Group Title

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

17. finaynay Group Title

for realsssss noooooooo

18. Callisto Group Title

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 Group Title

yeah

20. Callisto Group Title

So... just differentiate that guy :|

21. finaynay Group Title

so the derivative is 2x

22. finaynay Group Title

x=-1?

23. satellite73 Group Title

not quite

24. satellite73 Group Title

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

25. finaynay Group Title

how is it 9−x^2

26. satellite73 Group Title

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 Group Title

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

28. satellite73 Group Title

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 Group Title

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 Group Title

lets try it

31. satellite73 Group Title

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

32. finaynay Group Title

isn't it -2

33. satellite73 Group Title

34. finaynay Group Title

oh okay lol

35. math>philosophy Group Title

<3

36. satellite73 Group Title

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

37. finaynay Group Title

-2=f'(x)

38. finaynay Group Title

yessss! alright i get it now

39. satellite73 Group Title

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