Course:MATH102/Question Challenge/2004 December Q6

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Question

Consider the polynomial

$y = p (x) = K (x^{2} - 16) (x - a)$

where a,K are constants. Suppose that $p (0) = 12$ and that $p (x)$ has an inflection point at $x = 1 / 3$ .

(a) What are the values of the constants $x = a$ and $K$ ?

(b) Where does this polynomial have critical points?

(c) Which critical point is a maximum?

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Hint
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Solutions

Solution
$y = K (x^{2} - 16) (x - a)$ $K =$ an unknown constant $a =$ an unknown constant First let's expand the polynomial and ignore the constants to avoid using the product rule: $y = K (x^{3} - a x^{2} - 16 x + 16 a)$ Next we find the second derivative: $\frac{d y}{d x} = K (3 x^{2} - 2 a x - 16)$ $\frac{d^{2} y}{d x^{2}} = K (6 x - 2 a)$ We set the the second derivative equal to $0$ when $x = 1 / 3$ : $0 = K (6 (1 / 3) - 2 a)$ $0 = K (2 - 2 a)$ We can assume that $K$ does not equal zero as then the polynomial would not be very interesting if it was just the x-axis. Therefore: $2 = 2 a$ $a = 1$ We will substitute for the constant a and solve for the constant K: $12 = K ((0)^{3} - (1) (0)^{2} - 16 (0) + 16 (1)) = K (16)$ $K = \frac{12}{16} = \frac{3}{4}$ To find the critical points of the polynomial we will take the first derivative: $y = \frac{3}{4} (x^{3} - (1) x^{2} - 16 x + 16 (1))$ $y = \frac{3}{4} (x^{3} - x^{2} - 16 x + 16)$ $\frac{d y}{d x} = \frac{3}{4} (3 x^{2} - 2 x - 16)$ Setting $\frac{d y}{d x} = 0$ and solve for $x$ : $0 = \frac{3}{4} (3 x^{2} - 2 x - 16)$ $0 = 3 x^{2} - 2 x - 16$ This is a quadratic equation, so we will use the quadratic formula: $x = \frac{- (- 2) \pm (\sqrt{(- 2)^{2} - 4 \cdot 3 \cdot (- 16))}}{(2 \cdot 3)} = - 2, \frac{8}{3}$ To find which of these critical points is a local maximum we check the sign of the second derivative: $\frac{d y}{d x} = \frac{3}{4} (3 x^{2} + 2 x - 16)$ $\frac{d^{2} y}{d x^{2}} = \frac{3}{4} (6 x + 2)$ For $x = - 2$ $\frac{d^{2} y}{d x^{2}} = \frac{3}{4} (6 (- 2) + 2) = \frac{3}{4} (- 12 + 2) < 0$ For $x = \frac{8}{3}$ $\frac{d^{2} y}{d x^{2}} = \frac{3}{4} (6 (8 / 3) + 2) > 0$ When the second derivative is negative at $x = - 2$ , this is a local maximum. Part A) $a = 1$ and $K = 3 / 4$ Part B) The polynomial has critical points at $x = - 2$ and $x = 8 / 3$ Part C) The polynomial has a maximum at $x = - 2$ and a minimum at $x = 8 / 3$