Find the minimum and maximum values of f(x) = 1/(x²)+

Find the minimum and maximum values of $\displaystyle f(x) = \frac{1}{x^{2}}+\frac{1}{\bigl(1-x\bigr)^{2}}$.

Step 1: Domain and symmetry

The function is defined for $x\neq0$ and $x\neq1$, so the domain splits into $\bigl(-\infty,0\bigr)$, $\bigl(0,1\bigr)$ and $\bigl(1,+\infty\bigr)$. It is positive throughout, and symmetric about $x = \dfrac12$:

\[ f\bigl(1-x\bigr) = \frac{1}{\bigl(1-x\bigr)^{2}}+\frac{1}{x^{2}} = f(x). \]

That symmetry alone forces $x = \dfrac12$ to be a critical point.

Step 2: The derivative

\[ f'(x) = -\frac{2}{x^{3}}+\frac{2}{\bigl(1-x\bigr)^{3}} = \frac{2x^{3}-2\bigl(1-x\bigr)^{3}}{x^{3}\bigl(1-x\bigr)^{3}}. \tag{1} \]

Expanding the numerator: \[2x^{3}-2\bigl(1-3x+3x^{2}-x^{3}\bigr) = 4x^{3}-6x^{2}+6x-2 = 2\bigl(2x^{3}-3x^{2}+3x-1\bigr),\]

and since $x = \dfrac12$ is a root, $2x^{3}-3x^{2}+3x-1 = \bigl(2x-1\bigr)\bigl(x^{2}-x+1\bigr)$, where the quadratic has discriminant $1-4 = -3\lt 0$ and is therefore positive for every $x$. Hence

\[ f'(x) = \frac{2\bigl(2x-1\bigr)\bigl(x^{2}-x+1\bigr)}{x^{3}\bigl(1-x\bigr)^{3}}. \tag{2} \]

Step 3: Sign of the derivative

The factor $x^{2}-x+1$ is positive, so the sign of $f'$ is the sign of $\dfrac{2x-1}{x^{3}\bigl(1-x\bigr)^{3}}$: \[\begin{array}{c|c|c|c|c} \text{interval} & 2x-1 & x^{3} & \bigl(1-x\bigr)^{3} & f' \\ \hline \bigl(-\infty,0\bigr) & - & - & + & +\ (\nearrow) \\ \bigl(0,\tfrac12\bigr) & - & + & + & -\ (\searrow) \\ \bigl(\tfrac12,1\bigr) & + & + & + & +\ (\nearrow) \\ \bigl(1,+\infty\bigr) & + & + & - & -\ (\searrow) \end{array}\]

The only critical point is $x = \dfrac12$, and there $f'$ passes from $-$ to $+$: a local minimum, not a maximum.

Step 4: The critical value and the behaviour at the boundaries

\[ f\!\left(\frac12\right) = \frac{1}{\left(\tfrac12\right)^{2}}+\frac{1}{\left(\tfrac12\right)^{2}} = 4+4 = 8, \]

\[ \lim_{x\to0^{\pm}}f(x) = +\infty, \qquad \lim_{x\to1^{\pm}}f(x) = +\infty, \qquad \lim_{x\to\pm\infty}f(x) = 0^{+}. \]

Step 5: Assembling the picture

• On $\bigl(-\infty,0\bigr)$: increasing from $0^{+}$ to $+\infty$. No extremum.
• On $\bigl(0,\tfrac12\bigr)$: decreasing from $+\infty$ down to $8$.
• At $x = \tfrac12$: the minimum value $8$, which by symmetry is the least value on $\bigl(0,1\bigr)$.
• On $\bigl(\tfrac12,1\bigr)$: increasing from $8$ back up to $+\infty$.
• On $\bigl(1,+\infty\bigr)$: decreasing from $+\infty$ to $0^{+}$. No extremum.

Since $f\gt 0$ everywhere while $f\to0$ at infinity, the infimum over the whole domain is $0$ and is never attained; and $f$ is unbounded above. So there is no global minimum and no maximum at all — the only extremum anywhere is the local minimum $8$.

Step 6: The graph

Step 7: Verification

\[f(0.25) = 17.777778 \gt f(0.5) = 8 \lt f(0.75) = 17.777778 \;\;\checkmark\ \text{(a minimum)}\]

\[f'(0.25) = -123.2593\lt 0, \qquad f'(0.75) = +123.2593\gt 0 \;\;\checkmark\]

\[f''\!\left(\tfrac12\right) = 192\gt 0 \;\;\checkmark; \qquad f(-1) = f(2) = 1.25,\ f(-3) = 0.1736,\ f(10) = 0.0223 \;\;\checkmark\]