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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?
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?

This Question is Closed

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1if u plot it , it will be simpler to do

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[x^28=(x2\sqrt{2})(x+2\sqrt{2})\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436664302388:dw

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436664349288:dw

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436664378196:dw

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0So, 2 because those are the 0s?

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1can you tell me what kind of point , theorem want ?

zeesbrat3
 one year ago
Best ResponseYou've already chosen the best response.0You're looking for a point that is defined and continuous on the supplied interval as well as differentiable... right?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1HI!! (and holy moly!!0

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1you are not looking for a point that is defined and continuous, that doesn't even make any sense !!

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1first you have to check to see if your function is differentiable on the interval you are given

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436664956891:dw does it make sense ?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1since \[x^28=x^2+8\] on your interval, it is definitely continuous and differentiable (because it is a polynomial)

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1you are not asked do find any number here, just how many numbers will satisfy the conclusion of the mean value theorem

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1since your function is a quadratic, the derivative is a line when you solve a linear equation there is only one solution

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1in 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
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for your explanation!

misty1212
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.0If 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 $$ f'(c) = \frac{f(b)  f(a)}{ba} $$ 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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