Find the point(s) c guaranteed by the Mean Value Theorem for the function f(x)=(x+1)/x on the closed interval?

Find the point(s) c guaranteed by the Mean Value Theorem for the function f(x)=(x+1)/x on the closed interval [1/2, 2]





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Find the point(s) c guaranteed by the Mean Value Theorem for the function f(x)=(x+1)/x on the closed interval?
The Mean Value Theorem says





"If a function f is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in [a, b] such that f'(c) = (f(b) - f(a)) / (b - a)."





We want to find all numbers c that satisfy that equation above.





In this case, a = 1/2 and b = 2, and f(a) = f(1/2) = 3, and f(b) = f(2) = 3/2.





So the Mean Value Theorem guarantees that there's a c such that





f'(c) = (f(b) - f(a)) / (b - a) = (3/2 - 3) / (2 - 1/2) = (-3/2) / (3/2) = -1.





We must find all points in the interval [1/2, 2] where the derivative of f is -1. So let's differentiate f:





f(x) = (x + 1)/x, so


f'(x) = -1/(x^2)





For which x does this equal -1?





-1 = -1/(x^2)


1 = 1/(x^2)


x^2 = 1


x = -1, 1





Only x = 1 is in our interval, so this is the unique point where the derivative is -1.





Thus, c = 1 is the point which satisfies the Mean Value Theorem. (In some cases, there might be more than one value for c, but in this problem there is only one.)