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Difficulty: Very hardStationary Points, Maxima, and Minima

A cubic curve is defined by the equation y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d. The curve has a point of inflexion at (0,1)(0, 1) and a stationary point at (1,5)(1, 5). What is the local minimum value of the function?

  1. −3-3Answer
  2. B
    −7-7
  3. C
    55
  4. D
    −1-1

Answer

The local minimum value of the function is −3-3.
By applying the conditions for a point of inflexion (y′′=0y''=0) and a stationary point (y′=0y'=0), the curve equation is identified as y=−2x3+6x+1y = -2x^3 + 6x + 1. Solving y′=0y' = 0 gives x=−1x = -1 for the local minimum, resulting in a minimum value of −3-3.

Step-by-Step Solution

1
Use the point of inflexion and given point (0,1)(0, 1) to find bb and dd.
d=1d = 1 and b=0b = 0.
Since (0,1)(0,1) lies on the curve, substituting x=0,y=1x=0, y=1 gives d=1d = 1. The second derivative is y′′=6ax+2by'' = 6ax + 2b. At a point of inflexion x=0x = 0, y′′=0y'' = 0, giving 2b=0  ⟹  b=02b = 0 \implies b = 0.
2
Use the stationary point condition at (1,5)(1, 5) to determine aa and cc.
a=−2a = -2 and c=6c = 6.
The equation reduces to y=ax3+cx+1y = ax^3 + cx + 1. Since (1,5)(1,5) is on the curve, a+c+1=5  ⟹  a+c=4a + c + 1 = 5 \implies a + c = 4. Also, y′=3ax2+c=0y' = 3ax^2 + c = 0 at x=1x = 1, so 3a+c=03a + c = 0. Solving 3a+c=03a + c = 0 and a+c=4a + c = 4 simultaneously yields a=−2a = -2 and c=6c = 6.
3
Find all stationary points of y=−2x3+6x+1y = -2x^3 + 6x + 1 and test their nature using the second derivative.
Stationary points are at x=1x = 1 (maximum) and x=−1x = -1 (minimum).
Setting y′=−6x2+6=0y' = -6x^2 + 6 = 0 yields x2=1  ⟹  x=±1x^2 = 1 \implies x = \pm 1. The second derivative is y′′=−12xy'' = -12x. For x=−1x = -1, y′′=12>0y'' = 12 > 0, confirming a local minimum.
4
Evaluate the function at x=−1x = -1 to find the local minimum value.
y(−1)=−3y(-1) = -3.
Substituting x=−1x = -1 into y=−2x3+6x+1y = -2x^3 + 6x + 1 yields y=−2(−1)3+6(−1)+1=2−6+1=−3y = -2(-1)^3 + 6(-1) + 1 = 2 - 6 + 1 = -3.

Key Concept

Determining curve constants using stationary points and points of inflexion, followed by identifying local extrema.
Estimated Time:3m 0s
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Topics

Stationary Points, Maxima, and Minima

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