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If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$

I attempt:

I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....

But we can see:

$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$

Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).

Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?

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Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.

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  • $\begingroup$ should $f(4)=22$ $\endgroup$
    – mnulb
    Jul 11 '17 at 3:58
  • $\begingroup$ The OP wrote "with leading coefficient $1$". $\endgroup$ Jul 11 '17 at 3:59
  • $\begingroup$ I assumed that $f(6/5)=(6/5)^3$ to be proved from the given information that $f$ is monic. $\endgroup$
    – user348749
    Jul 11 '17 at 4:00
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Classic long method:

Let $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to \begin{align} f(1) &= 1 = 1 + b + c + d \hspace{10mm} \to d = -b - c \\ f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \hspace{10mm} \to c = -4 - 3b, \, d = 4 + 2b \\ f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b \end{align} from which $b = -5$, $c = 11$, and $d = -6$ and $$f(x) = x^3 - 5 \, x^2 + 11 \, x -6.$$

With $f(x)$ then \begin{align} f(4) &= 64 - 80 + 55 -6 = 22 \\ f\left(\frac{6}{5}\right) &= \left(\frac{6}{5}\right)^{3} - \frac{36 - 66 + 30}{5} = \left(\frac{6}{5}\right)^{3}. \end{align}

It may also be noticed that $f(x)$ can be seen in the form $$f(x) = \left(x - \frac{5}{3}\right)^{3} + \frac{8}{3} \, \left( x - \frac{5}{3}\right) + \frac{83}{27}.$$ From this it is easy to see that \begin{align} f\left(\frac{5}{3}\right) &= 3 + \frac{2}{27} \\ f\left(\frac{5}{6}\right) &= \frac{59}{216}. \end{align}

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    $\begingroup$ Looking at your last form of $f(x)$ if I put $5/3$ in then I would not get the same answer as you have for $f(5/3)$ just after - the former gives $3+2/27$... I've not chased the maths to see which is right, just happened to notice that discrepancy (or I've done something stupid which is also possible). $\endgroup$
    – Chris
    Jul 11 '17 at 8:47
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If

  • $f(x) = x^3 + ax^2 + bx + c$

  • $f(1) = 1$

  • $f(2) = 4$

  • $f(3) = 9$

then

\begin{align} f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\ \hline 4-1 &= f(2)-f(1) \\ 3 &= 7 + 3a + b \\ \hline 9-4 &= f(3) - f(2) \\ 5 &= 19 + 5a + b \\ \hline 3a + b &= -4 \\ 5a + b &= -14 \\ \hline a &= -5 \\ b &= 11 \\ c &= -6 \end{align}

S0 $f(x) = x^3 - 5x^2 + 11x - 6$.


You could make a difference table

\begin{array}{c} 1 && 4 && 9 \\ & 3 && 5 \\ && 2 \end{array}

Using $f(x) = x^3 + ax^2 + bx + c$, this corresponds to

\begin{array}{c} 1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\ & 7+3a+b && 19 + 5a + b \\ && 12+2a \end{array}

Then, comparing entries...

\begin{align} 12+2a=2 &\implies a = -5 \\ 7+3a+b = 3 &\implies b=11 \\ 1+a+b+c = 1 &\implies c = -6 \end{align}

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