f(x)=0. show that there is a root for equation: sin(x)=x^2-x for 1

Question

f(x)=0. show that there is a root for equation: sin(x)=x^2-x for 1<x<2

anonymous · Answer

We know $$-1\leq \sin x \leq 1$$ hence, $$-1\leq -\sin x \leq 1$$
Now since, $$1<x<2$$ and hence, $$-2<-x<-1$$
Therefore$$1<x^2<4$$

We can rewrite the equation as $$x^2-x-\sin x = 0$$
Therefore adding equations above we get $$-2<x^2-x-\sin x < 4$$
Therefore for some value of x the equation has a very equals to 0 and hence a root exists.

anonymous · Answer

Is this a question that you have in calculus?

anonymous · Answer

yes

anonymous · Answer

let f(x) = sinx - x^2 + x

a special case of the intermediate value theorem states:

if 0 is a number between f(1) and f(2), then there is a c in [1, 2] such that f(c) = 0 (ie: f has a root)

f(1) = sin(1) - 1 + 1 = sin(1), which is greater than 0

f(2) = sin(2) - 4 + 2 = sin(2) - 2, which is less than 0 

hence there is a c in [1,2] such that f(c) = 0

anonymous · Answer

technically the inverval should be (1,2) , ie not including the endpoints :P

anonymous · Answer

but if there is a c in [1,2] such that f(c) = 0, that implies there is a c in (1,2) since f(1) and f(2) aren't 0