I am stuck on this problem :

Use the intermediate value theorem to show that the equation sin(x)=x-1 has at least one solution in the interval [0,pi]

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- Apr 20th 2010, 07:17 PMsydewayzloccintermediate value theorem
I am stuck on this problem :

Use the intermediate value theorem to show that the equation sin(x)=x-1 has at least one solution in the interval [0,pi] - Apr 20th 2010, 07:21 PMDiemo
sin(x) =x-1

Rewrite this as x-1-sin(x)=0. Now, what value do you get when x=0, and what value do you get when x=pi, and what does the intermediate value theorem tell you in this case? - Apr 20th 2010, 07:24 PMsydewayzlocc
i get -1 and pi-1

- Apr 20th 2010, 07:30 PMDiemo
The important thing to note here is that pi-1 is greater than zero, wheras -1 is less than zero. The intermediate value theorem states that:

Suppose that f : [a, b] → R is continuous and that u is a real number satisfying f(a) < u < f(b) or f(a) > u > f(b). Then for some c ∈ [a, b], f(c) = u.

Ergo, there is at least one value in the range [0,pi] which satisfies sin(x)=1-x.

(As an aside, I always forget whether or not the square brackets or the ordinary brackets mean that those values are included in the domain, but the point remains the same, and it shouldn't make any difference provided that you use a value close to zero and close to pi).