# Show the equation f (x)-2x =0 have 1 solution <2 and >1. f co ntinue integral 1-2 f(x)dx =3

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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` .**

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=`

`G(x)= 3-[x^2/2]_1^2=3-(2^2-1^2)=0`

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`