Determine if the mean value theorem for integrals applies to the function f(x) = 3 - x^2 on the interval the closed interval from [0, √3]. if so, find the x-coordinates of the point(s) guaranteed by the theorem.

Answer is    x = 1======to use the mvt for integrals on the interval [a,b], we require that f is continuous on [a,b].for  f(x) = 3 - x², we have f as a polynomial function (it is a parabola). polynomial functions are continuous everywhere, so this is continuous on the closed inerval [0,  √3]. therefore, we can apply the mvt for integrals.the mvt for integrals says that there is some number c in the interval [a,b] such that    f(c) = average value of f on [a,b]so we need to find the average value of f first.the formula for average value of a function f continuous on [a,b] is    since here we have f(x) = 3 - x² and our interval is[0,√3]:     distribute 1 /  √3 into the parentheses.a cancellation will occur.also, (√3 )³ /  √3 = (√3)²    the average value of f on that interval is 2. applying mvt for integrals to find the x-coordinates guaranteed by theorem:     but x = -1 is not in the interval  [0, √3]. therefore, the only x-coordinate guaranteed by the theorem is      x = 1

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