Examkin

Question

Difficulty: Very hardStationary Points, Maxima, and Minima

A cubic curve defined by y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d has a local maximum at (−1,10)(-1, 10) and a point of inflexion at (1,2)(1, 2). What is the value of yy at the local minimum of the curve?

Answer: -6

Answer

The local minimum value of yy on the curve is −6-6.
By setting up a system of equations using the conditions for the local maximum at (−1,10)(-1, 10) and the point of inflexion at (1,2)(1, 2), the cubic curve is uniquely determined as y=0.5x3−1.5x2−4.5x+7.5y = 0.5x^3 - 1.5x^2 - 4.5x + 7.5. Setting the derivative dydx=1.5(x−3)(x+1)=0\frac{dy}{dx} = 1.5(x-3)(x+1) = 0 gives x=3x = 3 as the xx-coordinate of the local minimum. Evaluating y(3)y(3) gives −6-6.

Step-by-Step Solution

1
Differentiate the general cubic equation to obtain expressions for the first and second derivatives.
dydx=3ax2+2bx+c\frac{dy}{dx} = 3ax^2 + 2bx + c and d2ydx2=6ax+2b\frac{d^2y}{dx^2} = 6ax + 2b.
Stationary points require dydx=0\frac{dy}{dx} = 0 and points of inflexion require d2ydx2=0\frac{d^2y}{dx^2} = 0.
2
Apply the point of inflexion conditions at (1,2)(1, 2).
6a(1)+2b=0  ⟹  b=−3a6a(1) + 2b = 0 \implies b = -3a, and a(1)3+b(1)2+c(1)+d=2  ⟹  −2a+c+d=2a(1)^3 + b(1)^2 + c(1) + d = 2 \implies -2a + c + d = 2.
At a point of inflexion, the second derivative is zero, and the point lies on the curve.
3
Apply the stationary point and coordinate conditions at the local maximum (−1,10)(-1, 10).
3a(−1)2+2b(−1)+c=0  ⟹  9a+c=0  ⟹  c=−9a3a(-1)^2 + 2b(-1) + c = 0 \implies 9a + c = 0 \implies c = -9a, and −a+b−c+d=10  ⟹  5a+d=10-a + b - c + d = 10 \implies 5a + d = 10.
At a local maximum, the first derivative is zero, and the point lies on the curve.
4
Solve the system of linear equations for coefficients a,b,c,da, b, c, d.
a=0.5a = 0.5, b=−1.5b = -1.5, c=−4.5c = -4.5, d=7.5d = 7.5.
Combining −11a+d=2-11a + d = 2 and 5a+d=105a + d = 10 yields 16a=8  ⟹  a=0.516a = 8 \implies a = 0.5.
5
Find the xx-coordinate of the local minimum by solving dydx=0\frac{dy}{dx} = 0.
1.5x2−3x−4.5=0  ⟹  1.5(x−3)(x+1)=0  ⟹  x=31.5x^2 - 3x - 4.5 = 0 \implies 1.5(x - 3)(x + 1) = 0 \implies x = 3 (since x=−1x = -1 is the local maximum).
Evaluating d2ydx2(3)=6(0.5)(3)+2(−1.5)=6>0\frac{d^2y}{dx^2}(3) = 6(0.5)(3) + 2(-1.5) = 6 > 0 confirms a local minimum at x=3x = 3.
6
Calculate the value of yy at x=3x = 3.
y=0.5(3)3−1.5(3)2−4.5(3)+7.5=−6y = 0.5(3)^3 - 1.5(3)^2 - 4.5(3) + 7.5 = -6.
Substituting x=3x = 3 into the curve equation gives the value of yy at the local minimum.

Key Concept

Determining polynomial coefficients from stationary and inflexion point conditions to find extreme values.
Rate this question

Topics

Stationary Points, Maxima, and Minima

More questions from this topic