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Mean value theroem??
Show that [tex]x^2 = xsinx + cosx[/tex] is true only for two values of [tex]x \in {R}[/tex]
Intermediate value theorem
Mean value theorem (?)
I already know how to prove that there is al least one [tex]x \in [1,1.5][/tex] and another [tex]x \in [-1.5,-1][/tex] where the equation holds. The thing is that I'm not completely sure how to pove that they are unique, I have a geometric argument buy i feel it can be done using the mean value theorem.
Just for you to know, what i did to find out where the x's are, i took [tex]f(x) = x^2-xsinx-cosx[/tex] and gave values to the function it turns out that [tex]f(1) < 0[/tex] and [tex]f(1.5) > 0[/tex] so there must be at leat one [tex]x \in [1,1.5][/tex] where [tex]f(x) = 0[/tex] But that's it, I ran out of ideas although i feel I'm really close.
Any ideas or advices will be appreiciated
Homework Statement
Show that [tex]x^2 = xsinx + cosx[/tex] is true only for two values of [tex]x \in {R}[/tex]
Homework Equations
Intermediate value theorem
Mean value theorem (?)
The Attempt at a Solution
I already know how to prove that there is al least one [tex]x \in [1,1.5][/tex] and another [tex]x \in [-1.5,-1][/tex] where the equation holds. The thing is that I'm not completely sure how to pove that they are unique, I have a geometric argument buy i feel it can be done using the mean value theorem.
Just for you to know, what i did to find out where the x's are, i took [tex]f(x) = x^2-xsinx-cosx[/tex] and gave values to the function it turns out that [tex]f(1) < 0[/tex] and [tex]f(1.5) > 0[/tex] so there must be at leat one [tex]x \in [1,1.5][/tex] where [tex]f(x) = 0[/tex] But that's it, I ran out of ideas although i feel I'm really close.
Any ideas or advices will be appreiciated