Course:MATH102/Question Challenge/2004 December Q6

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Question

Consider the polynomial

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

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?

Hints

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: $y=K(x^{3}-ax^{2}-16x+16a)$ Next we find the second derivative: ${\frac {dy}{dx}}=K(3x^{2}-2ax-16)$ ${\frac {d^{2}y}{dx^{2}}}=K(6x-2a)$ We set the the second derivative equal to $0$ when $x=1/3$: $0=K(6(1/3)-2a)$ $0=K(2-2a)$ 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: $2=2a$ $a=1$ 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: $y={\frac {3}{4}}(x^{3}-(1)x^{2}-16x+16(1))$ $y={\frac {3}{4}}(x^{3}-x^{2}-16x+16)$ ${\frac {dy}{dx}}={\frac {3}{4}}(3x^{2}-2x-16)$ Setting ${\frac {dy}{dx}}=0$ and solve for $x$: $0={\frac {3}{4}}(3x^{2}-2x-16)$ $0=3x^{2}-2x-16$ 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: ${\frac {dy}{dx}}={\frac {3}{4}}(3x^{2}+2x-16)$ ${\frac {d^{2}y}{dx^{2}}}={\frac {3}{4}}(6x+2)$ For $x=-2$ ${\frac {d^{2}y}{dx^{2}}}={\frac {3}{4}}(6(-2)+2)={\frac {3}{4}}(-12+2)<0$ For $x={\frac {8}{3}}$ ${\frac {d^{2}y}{dx^{2}}}={\frac {3}{4}}(6(8/3)+2)>0$ 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 $x=8/3$