# 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?

Dec 17, 2017

$f \left(2\right) = 1.5$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} + \frac{1}{x}}{{x}^{2} - \frac{1 - \frac{1}{x} ^ 2}{\frac{1}{x} + \frac{1}{x} ^ 2}}$

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

$\textcolor{w h i t e}{f \left(x\right)} = \frac{{x}^{2} + \frac{1}{x}}{{x}^{2} - \left(x - 1\right)}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{{x}^{2} + \frac{1}{x}}{{x}^{2} - x + 1}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{{x}^{3} + 1}{x \left({x}^{2} - x + 1\right)}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{\left(x + 1\right) \left({x}^{2} - x + 1\right)}{x \left({x}^{2} - x + 1\right)}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{x + 1}{x}$

$\textcolor{w h i t e}{f \left(x\right)} = 1 + \frac{1}{x}$

This function is monotonically decreasing over the interval $\left[1 , 2\right]$

So the minimum value occurs when $x = 2$, namely $f \left(2\right) = 1 + \frac{1}{2} = \frac{3}{2} = 1.5$