zeesbrat3
  • zeesbrat3
If f(x) = ∣(x2 − 8)∣, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem?
Mathematics
schrodinger
  • schrodinger
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zeesbrat3
  • zeesbrat3
Please help
amoodarya
  • amoodarya
if u plot it , it will be simpler to do
amoodarya
  • amoodarya
\[|x^2-8|=|(x-2\sqrt{2})(x+2\sqrt{2})|\]

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amoodarya
  • amoodarya
|dw:1436664302388:dw|
amoodarya
  • amoodarya
|dw:1436664349288:dw|
amoodarya
  • amoodarya
|dw:1436664378196:dw|
zeesbrat3
  • zeesbrat3
So, 2 because those are the 0s?
zeesbrat3
  • zeesbrat3
amoodarya
  • amoodarya
can you tell me what kind of point , theorem want ?
zeesbrat3
  • zeesbrat3
You're looking for a point that is defined and continuous on the supplied interval as well as differentiable... right?
amoodarya
  • amoodarya
yes
misty1212
  • misty1212
HI!! (and holy moly!!0
misty1212
  • misty1212
you are not looking for a point that is defined and continuous, that doesn't even make any sense !!
misty1212
  • misty1212
first you have to check to see if your function is differentiable on the interval you are given
amoodarya
  • amoodarya
|dw:1436664956891:dw| does it make sense ?
misty1212
  • misty1212
since \[|x^2-8|=-x^2+8\] on your interval, it is definitely continuous and differentiable (because it is a polynomial)
misty1212
  • misty1212
you are not asked do find any number here, just how many numbers will satisfy the conclusion of the mean value theorem
misty1212
  • misty1212
since your function is a quadratic, the derivative is a line when you solve a linear equation there is only one solution
misty1212
  • 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
zeesbrat3
  • zeesbrat3
Thank you for your explanation!
misty1212
  • 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
ybarrap
  • 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

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