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Lemma
Stationary Points Difficulty 4

TMUA practice: Stationary Points, problem 3

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

The question
For which value of the real constant cc does the curve   y=x3−3x+c  \;y = x^{3} - 3 x + c\; have its local minimum value equal to 00?
A c=0c = 0
B c=−2c = -2
C c=1c = 1
D c=3c = 3
E c=2c = 2

The correct answer is highlighted. E

Worked solution

Locate the stationary points. Differentiate:   y′=3x2−3=3(x−1)(x+1)\;y' = 3 x^{2} - 3 = 3(x - 1)(x + 1). Solve y′=0y' = 0: x=1x = 1 or x=−1x = -1.

Classify. Second derivative:   y′′=6x\;y'' = 6 x.

  • At x=1x = 1:   y′′(1)=6>0\;y''(1) = 6 > 0. Local minimum. Minimum value:   y(1)=1−3+c=c−2\;y(1) = 1 - 3 + c = c - 2.
  • At x=−1x = -1:   y′′(−1)=−6<0\;y''(-1) = -6 < 0. Local maximum. Maximum value:   y(−1)=−1+3+c=c+2\;y(-1) = -1 + 3 + c = c + 2.

Impose the condition. The local minimum value equals 00:

c−2  =  0  ⟹  c  =  2.c - 2 \;=\; 0 \;\Longrightarrow\; c \;=\; 2.

This is option E.

Verify: with c=2c = 2,   y=x3−3x+2\;y = x^{3} - 3 x + 2. Factor: at x=1x = 1, y(1)=1−3+2=0y(1) = 1 - 3 + 2 = 0 ✓. (In fact y=(x−1)2(x+2)y = (x - 1)^{2}(x + 2), so x=1x = 1 is a double root — the curve touches the xx-axis at the local minimum.)

Why the other options fail.

  • A, c=0c = 0:   y(1)=−2\;y(1) = -2, not 00.
  • B, c=−2c = -2:   y(1)=−4\;y(1) = -4, not 00.
  • C, c=1c = 1:   y(1)=−1\;y(1) = -1.
  • D, c=3c = 3:   y(1)=1\;y(1) = 1.

The lesson: ‘local minimum equals 0’ means the yy-value at the local minimum is zero. Find the local-min xx first, then yy at that xx, then solve for the parameter.