Mathematico | Mathematico

Let

y = \frac{1 - 6 x + 3 x^{2}}{x^{2}}, x \neq 0

Find the smallest possible value of $y$ .

Hint

Split this into $3$ fractions:

y = \frac{1}{x^{2}} - \frac{6 x}{x^{2}} + \frac{3 x^{2}}{x^{2}}

Hint

After simplifying you get

y = \frac{1}{x^{2}} - \frac{6}{x} + 3

This is a quadratic in disguise!

Hint

We have

y = {(\frac{1}{x})}^{2} - 6 (\frac{1}{x}) + 3

Let $u = \frac{1}{x}$ and complete the square to find the minimum value.

Solution

We split this into three fractions and then simplify

\begin{aligned} y & = \frac{1}{x^{2}} - \frac{6 x}{x^{2}} + \frac{3 x^{2}}{x^{2}} \\ = \frac{1}{x^{2}} - \frac{6}{x} + 3 \end{aligned}

Now let $u = \frac{1}{x}$ and complete the square

\begin{aligned} y & = u^{2} - 6 u + 3 \\ = (u - 3)^{2} - 9 + 3 \\ = (u - 3)^{2} - 6 \end{aligned}

The minimum value is $6$ and occurs when $u = 3 \Rightarrow x = \frac{1}{3}$ .