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Show the equation f (x)-2x =0 have 1 solution <2 and >1. f co ntinue integral...
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You need to consider the following function `g(x) = f(x) - 2x` , hence, integrating the function `g(x)` yields:
`int_1^2 g(x) dx = int_1^2 (f(x) - 2x) dx`
Using the proeprty of linearity of integral yields:
`int_1^2 g(x) dx = int_1^2 (f(x))dx - int_1^2 2x dx`
Since the problem provides the information that `int_1^2 (f(x))dx = 3` , yields:
`int_1^2 g(x) dx = 3 - 2 int_1^2 x dx`
`int_1^2 g(x) dx = 3 - 2 x^2/2|_1^2`
Reducing duplicate factros and using the fundamental theorem of calcululs yields:
`int_1^2 g(x) dx = 3 - (2^2 - 1^2) = 0`
Since the problem provides the information that `f(x)` is continuous over `[1,2]` , hence, g(x) is also continuous over `[1,2]` , thus, you may use the mean value theorem, such that:
`g(c) = (int_(1^2)g(x)dx)/(2-1)`
Replacing c for x in equation of `g(x) ` yields:
`f(c) - 2c = 0=> c in [1,2]` is the solution of the equation `f(x) - 2x = 0`
Hence, testing if there exists a solution to equation `f(x) - 2x = 0, x in [1,2]` , yields, using the mean value theorem, that there exists `c in [1,2],` such that `f(c) - 2c = 0` .
Posted by sciencesolve on May 19, 2013 at 5:04 PM (Answer #1)
Then we can say:
`G(x)= int( f(x)-2) dx`
`G(1)-G(2)=int_1^2 (f(x)-2x) dx` `=int_1^2 f(x) dx- int_1^2 2xdx=`
Since `f(x)` is continue `f(x)-2x` is also contiue, so for Torricelli Barrow Theorem, `G(x)` , is the integral function of `f(x)-2x` , and also continue.
So: `G(x) ` has the same value for `x=1` and `x=2`
Then for Lagranges Theorem: `EE xi , 1 <= xi <= 2` , so that:
`G'( xi)(2-1)= 0` `rArr` `G'(xi)=0` `rArr` `f(xi)-2xi=0`
That means `xi` is solution of `f(x)-2x=0`
Posted by oldnick on May 20, 2013 at 2:25 AM (Answer #2)
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