To show such a solution exists, we almost always employ the IVT. It is easiest to use the difference of functions, in other words when f(x) = g(x) then f(x) - g(x) = 0. Then, often we can simply plug in the endpoints of the interval to show that f - g changes signs:
f-g(x) = sin(x2 + 3x) + xcosx continuous because both f and g are continuous and the IVT applies
f-g(-π) = sin(π2- 3π) + π > 0 (since -1 ≤ sinx ≤ 1)
f-g(-π/6) = sin(π2/36 - π/2) - π√3/12 < 0 (by calculator)
f-g(-π/6) < 0 < f-g(-π)
So, by the IVT, there exists an x0 with -π < x0 < -π/6 such that f(x0) - g(x0) = 0 and f(x0) = g(x0)
Josh F.
tutor
Yes, exactly right. That is the x-coordinate of the point of intersection of the two functions, which is guaranteed to occur in the given x-interval, - pi to - pi/6. You could go to desmos.com and graph the two to confirm -- or graph the difference of the two to confirm it has a zero on the interval.
Glad it helped!
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04/01/21
Maggie D.
This is insanely helpful and I can't thank you enough. At the end, when you say X0, does that stand for the meeting points of the functions?04/01/21