Weekly Challenge 27: Cubic roots

7066 /www/nrich/html/content/id/7066/ 1 2011-03-15T09:38:41 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="buttonBar"> <li><a href="http://nrich.maths.org/791&part=">Warm-up problem</a></li> <li><a href="http://nrich.maths.org/6428&part=">Try this next</a></li> <li><a href="https://nrich.maths.org/discus/messages/27/27.html">Ask NRICH</a></li> <li><a href="http://nrich.maths.org/1422&part=">Read all about it</a></li> <li><a href="http://nrich.maths.org/7058&part=http://nrich.maths.org/7058&part=solution">Last week's solution</a></li> </ul> <div> A certain cubic polynomial $y=f(x)$ cuts the $x$-axis at the three points $x=10, 100$ and $1000$. Is this enough information to determine the location of its point of inflection (note that this is not necessarily a stationary point of inflection)? If so, where is this point; if not, why not? Construct a cubic polynomial which cuts the $x$-axis at $x=10, 100$ and its point of inflection. How many such polynomials are there? </div> <div class="framework">Did you know ... ? Polynomials have many fascinating properties. A key result of university mathematics is the Fundamental Theorem of Algebra which states that any polynomial of degree $n$ $p(z)= a_nz^n+a_{n-1}z^{n-1}+\dots+a_0$, with $a_n\neq 0$, has precisely $n$, possibly repeated, complex number solutions.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> A polynomial $f(x)$ has a factor $(x-a)$ if and only if $f(a)=0$. Thus, a polynomial cutting the $x$-axis at $10, 100, 1000$ has factors $(x-10)(x-100)(x-1000)$. This defines a cubic polynomial up to a multiplicative factor. Thus $$f(x) = A(x-10)(x-100)(x-1000) = A\left(x^3 -(10+100+1000)x^2 + .. \right)\,,$$ for some constant $A$. Now, a point of inflection necessarily has $f''(x) = 0$. Only the $x^3$ and $x^2$ terms of a cubic polynomial contributes to its second derivative, so there is no need to expand the polynomial in full to see that $$ f''(x) = 6Ax-2220A $$ This is zero at the single point $x = \frac{2220}{6} = 370$. Therefore the point of inflection for the cubic is at $x=370$, regardless of the choice of $A$. For the second part, the polynomial must take the form $$ f(x) = A(x-10)(x-100)(x-a)\quad \mbox{for a constant } a \mbox{ where} \quad f''(a)=0 $$ So, we need to take the second derivative to work out the constraints on $a$. I will keep the form of the factors and use the chain rule to make life simple, although you could expand the brackets first if you wish $$ f''(x) = 2A\left((x-10)+(x-100)+(x-a)\right) $$ So, $$ f''(a) = 2A(2a-110)=0 $$ Since $A$ cannot be zero for a cubic polynomial, we must have $a=55$. The polynomial must therefore be of the form $$ f(x) = A(x-10)(x-55)(x-100) $$ Alternative, quick, method for second part: From the first part of the question I noticed a generalisation that the point of inflection of a cubic is found at one third of the sum of the roots $r_1+r_2+r_3$. If one of the roots is the point of inflection then $$ r_1+a+r_2 = 3a $$ Thus, the point of inflection which is a root is found at one-half of the sum of the other two roots. Thus, in our special case, $$ a = \frac{1}{2}(10+100) = 55, \mbox{ as before}. $$ Isn't maths great! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> A polynomial $f(x)$ has a factor $(x-a)$ if and only if $f(a)=0$. Thus, a polynomial cutting the $x$-axis at $10, 100, 1000$ has factors $(x-10)(x-100)(x-1000)$. This defines a cubic polynomial up to a multiplicative factor. Thus $$f(x) = A(x-10)(x-100)(x-1000) = A\left(x^3 -(10+100+1000)x^2 + .. \right)\,,$$ for some constant $A$. Now, a point of inflection necessarily has $f''(x) = 0$. Only the $x^3$ and $x^2$ terms of a cubic polynomial contributes to its second derivative, so there is no need to expand the polynomial in full to see that $$ f''(x) = 6Ax-2220A $$ This is zero at the single point $x = \frac{2220}{6} = 370$. Therefore the point of inflection for the cubic is at $x=370$, regardless of the choice of $A$. For the second part, the polynomial must take the form $$ f(x) = A(x-10)(x-100)(x-a)\quad \mbox{for a constant } a \mbox{ where} \quad f''(a)=0 $$ So, we need to take the second derivative to work out the constraints on $a$. I will keep the form of the factors and use the chain rule to make life simple, although you could expand the brackets first if you wish $$ f''(x) = 2A\left((x-10)+(x-100)+(x-a)\right) $$ So, $$ f''(a) = 2A(2a-110)=0 $$ Since $A$ cannot be zero for a cubic polynomial, we must have $a=55$. The polynomial must therefore be of the form $$ f(x) = A(x-10)(x-55)(x-100) $$ Alternative, quick, method for second part: From the first part of the question I noticed a generalisation that the point of inflection of a cubic is found at one third of the sum of the roots $r_1+r_2+r_3$. If one of the roots is the point of inflection then $$ r_1+a+r_2 = 3a $$ Thus, the point of inflection which is a root is found at one-half of the sum of the other two roots. Thus, in our special case, $$ a = \frac{1}{2}(10+100) = 55, \mbox{ as before}. $$ Isn't maths great! </mdoxml> 2 3 0 0 0 0 1 Weekly Challenge 27: Cubic roots Find the location of the point of inflection of this cubic. Collections Weekly Challenge Admin Stage 5 - Core Mapping Stage 5 Core Mapping Document Algebra and Functions AS Secondary Mapping Document DisplayCabinet