Calculus 1 intermediate value theorem.

Question

Use intermediate value theorem to show there is a root of sinx-x^2+x=0 on interval (1,2)?

Answer 1

Let f(x) = sin x - x^2 + x

f(1) >0 and f(2)<0

IVT guarantees c in (1,2) such that f(c)=0

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Now, are you required to PROVE that f(1)>0 and f(2)<0 ????

f(1) = sin 1 - 1^2 + 1 = sin 1 -1+1 = sin 1 is in quadrant 1 where sine is positive

f(2) = sin 2 - 2^2 + 2 = sin 2 - 2; this is the 2nd quadrant where sine is positive;

sin(pi/2) = 1 and sin(pi) = 0, so 0 < sin t < 1 for pi/2 < t < pi

subtracting 2 from this inequality chain shows that the function is negative there