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Lemma
Stationary Points Difficulty 4

TMUA practice: Stationary Points, problem 1

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

The question
The curve   y=3x5−5x3  \;y = 3x^{5} - 5x^{3}\; has stationary points at x=1x = 1 and x=−1x = -1. Which one of the following correctly classifies them?
A x=1 is a local maximum, x=−1 is a local minimumx = 1 \text{ is a local maximum, } x = -1 \text{ is a local minimum}
B x=1 is a local minimum, x=−1 is a local maximumx = 1 \text{ is a local minimum, } x = -1 \text{ is a local maximum}
C both are local minima\text{both are local minima}
D both are local maxima\text{both are local maxima}
E both are horizontal points of inflection\text{both are horizontal points of inflection}

The correct answer is highlighted. B

Worked solution

Apply the second-derivative test. We have

y′′(x)  =  60x3−30x  =  30x(2x2−1).y''(x) \;=\; 60 x^{3} - 30 x \;=\; 30 x (2 x^{2} - 1).

At x=1x = 1:   y′′(1)=30(1)(2−1)=30>0\;y''(1) = 30(1)(2 - 1) = 30 > 0. Local minimum. Value y(1)=3−5=−2y(1) = 3 - 5 = -2; point (1, −2)(1,\, -2).

At x=−1x = -1:   y′′(−1)=30(−1)(2−1)=−30<0\;y''(-1) = 30(-1)(2 - 1) = -30 < 0. Local maximum. Value y(−1)=−3+5=2y(-1) = -3 + 5 = 2; point (−1, 2)(-1,\, 2).

So x=1x = 1 is a local minimum and x=−1x = -1 is a local maximum. This is option B.

Why the other options fail.

  • A: swaps the two classifications.
  • C, both minima: would require y′′y'' positive at both points; y′′(−1)=−30<0y''(-1) = -30 < 0.
  • D, both maxima: would require y′′y'' negative at both; y′′(1)=+30>0y''(1) = +30 > 0.
  • E, both inflection: would require y′′(x0)=0y''(x_{0}) = 0 at both points (and a sign-change check). Here y′′≠0y'' \ne 0 at ±1\pm 1.

The lesson: for an odd polynomial like 3x5−5x33x^{5} - 5x^{3}, the relative max sits at negative xx (the graph is decreasing for large negative xx — wait, here actually it descends from +∞+\infty on the left, reaches a max at −1-1, descends through 00 with a horizontal pause, reaches a min at +1+1, then ascends to +∞+\infty). The symmetry f(−x)=−f(x)f(-x) = -f(x) ensures the max and min are reflections of each other.