# 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 1 le x le 2. Write answer as *exact* decimal?

Jan 6, 2018

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

#### Explanation:

Given:

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)

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

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

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

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

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

with excluded value $x \ne - 1$

Note that $\frac{1}{x}$ is monotonically decreasing in the interval $\left[1 , 2\right]$.

So the minimum value is attained when $x = 2$:

$f \left(2\right) = 1 + \frac{1}{2} = 1.5$

Jan 6, 2018

$\frac{3}{2}$

#### Explanation:

$\text{Multiply numerator and denominator by (1/x + 1/x²).}$
$\text{Then we obtain f(x) = : }$

((x^2 + 1/x)(1/x + 1/x^2)) / ((x^2(1/x + 1/x^2) + 1/x^2 - 1)
$= \frac{x + 1 + \frac{1}{x} ^ 2 + \frac{1}{x} ^ 3}{x + 1 + \frac{1}{x} ^ 2 - 1}$

$\text{Now multiply numerator and denominator by x³ : }$

$= \frac{{x}^{4} + {x}^{3} + x + 1}{{x}^{4} + x}$

$\text{Now derive using the quotient rule.}$

$f ' \left(x\right) = \frac{\left(4 {x}^{3} + 3 {x}^{2} + 1\right) \left({x}^{4} + x\right) - \left({x}^{4} + {x}^{3} + x + 1\right) \left(4 {x}^{3} + 1\right)}{{x}^{4} + x} ^ 2$

$\text{The derivative is zero when the numerator is zero :}$

$- {x}^{6} - 2 {x}^{3} - 1 = 0$
$\implies {\left({x}^{3} + 1\right)}^{2} = 0$
$\implies {x}^{3} + 1 = 0$
$\implies x = - 1$

$\text{There is a problem though as the denominator is also zero}$
$\text{for this value of x, so we have the case 0/0, so we must}$
$\text{divide away the common factor (x+1)^2 :}$

$f ' \left(x\right) = - {\left({x}^{3} + 1\right)}^{2} / \left({x}^{2} {\left({x}^{3} + 1\right)}^{2}\right)$
$= - \frac{1}{x} ^ 2$

$\text{So the derivative is always negative. This means that the}$
$\text{function is ever decreasing, so the minimum over the interval}$
$\text{[1, 2] is reached for x=2 :}$

$f \left(2\right) = \frac{{2}^{4} + {2}^{3} + 2 + 1}{{2}^{4} + 2}$
$= \frac{27}{18}$
$= \frac{3}{2}$

Jan 6, 2018

Minimum value is $1.5$

#### Explanation:

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

= ((x^3+1)/x)/(x^2-(x^2(1-1/x^2))/(x^2(1/x+1/x^2))

= ((x^3+1)/x)/(x^2-(x^2-1)/(x+1)

= ((x^3+1)/x)/(x^2-(x-1)

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

= $\frac{x + 1}{x} = 1 + \frac{1}{x}$

Observe that at $x = 1$ we have $f \left(x\right) = 2$ and as $x$ increases to $2$, the value of $\frac{1}{x}$ comes down

and it is minimum wheen $x = 2$ and it is $1 + \frac{1}{2} = 1.5$

graph{(x^2+1/x)/(x^2-(1-1/x^2)/(1/x+1/x^2) [-0.983, 4.017, 0.23, 2.73]}