The problem to use the intermediate value theorem to show that there is a root of the given equation in the specified interval.

sin(x)= x^2-x, (1,2)

Can someone see if this is correct?

f(x) = sin(x) - x^2 + x, then f is continuous since sin(x) and x^2 - x are continuous and their composition is also continuous. sin(x) = x^2 - x which is equivalent to showing that f(x) = 0. f(1) = 0.841 and f(2) = -1.091, 0 s lying between f(1) and f(2). Since f is continuous, there is a c in (1,2) such that f(c) = 0
there is a root to the equation
sin(x) = x^2 - x

Question

The problem to use the intermediate value theorem to show that there is a root of the given equation in the specified interval.
 
sin(x)= x^2-x, (1,2)

Can someone see if this is correct?

f(x) = sin(x) - x^2 + x, then f is continuous since sin(x) and x^2 - x are continuous and their composition is also continuous. sin(x) = x^2 - x which is equivalent to showing that f(x) = 0.  f(1) = 0.841 and f(2) = -1.091, 0 s lying between f(1) and f(2). Since f is continuous, there is a c in (1,2) such that f(c) = 0
 there is a root to the equation
 sin(x) = x^2 - x

anonymous · Answer

@jim_thompson5910 can you check my work for me? :)

anonymous · Answer

I'm sure it is right, but I'm not this jim.

anonymous · Answer

LOL! okay :) is there a specific format the answer needs to be in?

zarkon · Answer

I wouldn't use 

'their composition'

zarkon · Answer

'their difference'

would be better

jim_thompson5910 · Answer

You are correct, all polynomials are continuous

Sine is continuous, so sin(x) - x^2 + x is continuous

f(1) = 0.841 and f(2) = -1.091

So because of the sign change and because f(x) is continuous, there is a number c such that f(c) = 0 since f(1) < 0 < f(2)

anonymous · Answer

Thank you guys!