Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. sin⁡x=x2−x,(1,2)

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Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.  sin⁡x=x2−x,(1,2)

broliY · Accepted Answer

Given expression is sin⁡x=x2−x for (1,2) Simplifying further ,we get x2−x−sin⁡x=0 As we know that polynomials and sine function are always continuous hence the above addition of polynomials and sine expression will be continuous in the given interval (1,2) Let f(x)=x2−x−sin⁡x where interval endpoints are 1 and 2 In radians it is between 0 and π thus there is positive value between 0 and 1 cheking for end points, we get at x=1 f(1)=12−1−sin⁡1 =1−1−sin⁡57.3∘ ≈−0.84&amp;lt;0 Since 1 radian =57.3∘ and sin⁡57.3∘≈sin⁡60∘ Similarly for x=2 f(2)=22−2−sin⁡2 =2−sin⁡114.6∘ ≈1.1&amp;gt;0 According to intermediate value theorem continuous function always take every value at least once between one point of graph to another point. Thus 0 is between −0.84 and 1.1 therefore ,the given equation sin⁡x=x2−x will have at least one root in (1,2).

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)

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