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Lemma
Stationary Points Difficulty 4

TMUA practice: Stationary Points, problem 2

A TMUA-calibre stationary points problem at difficulty 4 of 5, with the full worked solution.

The question
What is the unique stationary point of the curve   y=x2+1x2−1  \;y = \dfrac{x^{2} + 1}{x^{2} - 1}\; on its domain?
A (1, 0)(1,\, 0)
B (−1, 0)(-1,\, 0)
C (0, −1)(0,\, -1)
D (0, 1)(0,\, 1)
E (1, −1)(1,\, -1)

The correct answer is highlighted. C

Worked solution

Find y′y' by the quotient rule. Let u=x2+1u = x^{2} + 1 and v=x2−1v = x^{2} - 1. Then u′=2xu' = 2x and v′=2xv' = 2x, so

y′  =  u′v−uv′v2  =  2x(x2−1)−(x2+1)(2x)(x2−1)2  =  2x [(x2−1)−(x2+1)](x2−1)2  =  −4x(x2−1)2.y' \;=\; \frac{u' v - u v'}{v^{2}} \;=\; \frac{2x(x^{2} - 1) - (x^{2} + 1)(2x)}{(x^{2} - 1)^{2}} \;=\; \frac{2x\,\bigl[(x^{2} - 1) - (x^{2} + 1)\bigr]}{(x^{2} - 1)^{2}} \;=\; \frac{-4x}{(x^{2} - 1)^{2}}.

Solve y′=0y' = 0. The denominator (x2−1)2(x^{2} - 1)^{2} is positive on the domain x≠±1x \ne \pm 1. The numerator −4x=0-4x = 0 gives x=0x = 0. So the unique stationary point of yy on its domain is at x=0x = 0.

Compute y(0)y(0):   y(0)=(0+1)/(0−1)=1/(−1)=−1\;y(0) = (0 + 1)/(0 - 1) = 1/(-1) = -1.

The stationary point is (0, −1)(0,\, -1). This is option C.

(Note: x=±1x = \pm 1 are singularities of yy — the function is undefined there, and the graph has vertical asymptotes. These are not stationary points.)

Why the other options fail.

  • A, (1, 0)(1,\, 0) and B, (−1, 0)(-1,\, 0): x=±1x = \pm 1 are not in the domain of yy; the function is undefined there (denominator zero).
  • D, (0, 1)(0,\, 1): sign error — y(0)=−1y(0) = -1, not +1+1.
  • E, (1, −1)(1,\, -1): combines a singular xx with the correct yy-value at the stationary point, not a coherent answer.

The lesson: the quotient rule for y=u/vy = u/v is   y′=(u′v−uv′)/v2\;y' = (u'v - uv')/v^{2}. Watch the sign of the numerator — here the algebraic simplification leaves −4x-4x, not +4x+4x.