Show that the following equation have at least one solution in the
given intervals.
x cos x - 2x^2+ 3x-1=0, [0.2, 0.3] and [1.2,1.3]

Hi foxxybrown,
We can use the Bolzano Theorem to show that the given equation has at
least one solution in each of the given intervals.

Bolzano Theorem: "Let, for two real a and b, a < b, a function f be
continuous on a closed interval [a, b] such that f(a) and f(b) are of
opposite signs. Then there exists a number x0 in [a, b] with f(x0)=0."

http://www.cut-the-knot.org/Generalization/ivt.shtml
(proof of this thorem can be found following a link at the bottom of this page)

When we evaluate your equation in each of the extremes of the interval we get:

0.2cos(0.2) - 2(0.2)^2 + 3(0.2) - 1 = -0.28...
0.3cos(0.2) - 2(0.3)^2 + 3(0.3) - 1 = 0.0066...

Therefore, using the theorem, there exists a value between 0.2 and 0.3
such that x cos x - 2x^2+ 3x-1=0

Also:
1.2cos(1.2) - 2(1.2)^2 + 3(1.2) - 1 = 0.1548...
1.3cos(1.2) - 2(1.3)^2 + 3(1.3) - 1 = -0.132...

Therefore, there also exists a value between 1.2 and 1.3 such that x
cos x - 2x^2+ 3x-1=0


Google search terms:
bolzano theorem


I hope this helps!

Best wishes,
elmarto