Find the minimum value of f(x)=(x^2+1/x)/(x^2-(1-1/x^2)/(1/x+1/x^2) over the interval 1lexle2. Write your answer as an exact decimal?

1 Answer
Dec 17, 2017

f(2) = 1.5

Explanation:

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

color(white)(f(x)) = (x^2+1/x)/(x^2-(x^2-1)/(x+1)

color(white)(f(x)) = (x^2+1/x)/(x^2-(x-1))

color(white)(f(x)) = (x^2+1/x)/(x^2-x+1)

color(white)(f(x)) = (x^3+1)/(x(x^2-x+1))

color(white)(f(x)) = ((x+1)(x^2-x+1))/(x(x^2-x+1))

color(white)(f(x)) = (x+1)/x

color(white)(f(x)) = 1+1/x

This function is monotonically decreasing over the interval [1, 2]

So the minimum value occurs when x=2, namely f(2) = 1+1/2 = 3/2 = 1.5