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$P(x)$ is a $6^{\mathrm{th}}$ degree polynomial with coefficient of $x^6$ equal to $1$. And:

$P(1)=7$,

$P(2)=10$,

$P(3)=13$,

$P(4)=16$,

$P(5)=19$,

I have to find $P(6)$?

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it will be useful if you can show your work –  felasfa Aug 14 '13 at 8:29
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What is $f(x) - 3x - 4$ for the given values? –  Arthur Aug 14 '13 at 8:29
    
I tried using Lagrange intepolation, but since this method gives the polynomial with smallest degrees that satisfy the condition it gave me a result of: $$f(x) = 4 + 3x\text{, which is true}$$ You may try another type of intepolation. –  Stefan4024 Aug 14 '13 at 10:29

5 Answers 5

A monic polynomial of degree $n$ is uniquely determined by $n$ function values.

In your case, only $5$ such values ($f(1)$, $f(2)$, $f(3)$, $f(4)$ and $f(5)$) are given, meaning that $f(6)$ can be anything. For every choice of $f(6)$, there will be a suitable monic polynomial of degree $6$ by the above property.

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Just to get you started

In this case the polynomial can be represented by" $$ f(x)=x^{6}+a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x^{1}+a_{0} $$

Now you have 6 unknowns and 5 equations. It seems you are missing an extra information. If so, solve the system of equations to find the coefficients. Then you have the polynomial and can just evaluate $f(6)$.

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Aren't there $6$ unknowns? –  Alraxite Aug 14 '13 at 8:31
    
@Alraxite, you are right, for some reason I did forget to count the constant term. –  felasfa Aug 14 '13 at 8:36

As @Stefan4024 has remarked the first degree polynomial $p_0(x):=3x+4$ realizes the given values. The polynomial $q(x):=P(x)-p_0(x)$ should therefore be of degree $6$, have highest coefficient $1$, and should vanish at $1$,$2$, $\ldots$, $5$. Such a $q$ is necessarily of the form $$q(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-c)$$ for some $c\in{\mathbb R}$ (or $\in{\mathbb C}$). It follows that the most general $P$ satisfying your conditions is given by $$P(x)=3x+4+(x-1)(x-2)(x-3)(x-4)(x-5)(x-c)\ .$$ We now compute $$P(6)=18+4+5! (6-c)=742-120 c$$ and can immediately verify that the value $P(6)$ can yet be anything.

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plz see what is wrong in my method? –  Ankesh Aug 15 '13 at 10:44
    
@Ankesh: Your $h$ is not the most general polynomial of degree $6$ having the five given zeros, because you forced the sixth zero to be at $0$, not at a general $c$. –  Christian Blatter Aug 15 '13 at 10:56
    
okay i got your point. but if we get to know the f(0) value then my solution is write as we can take x instead of x-c? am i right? –  Ankesh Aug 15 '13 at 11:25

I tried like this only: $$ 742. $$ Explanation: Since the function value is in A.P. (i.e $7,10,13,16,19$), the $n$-th term is: $$ 7+(n-1)3=3n+4 $$ Leading coefficient is given as $1$. Now let polynomial be $h(x)=f(x)-(3x+4)$ which is also of degree $6$.

So that $h(x)$ will always be $0$ for $x=1$ to $5$. So $1,2,3,4,5$ are factors of $f(x)$. So it is roots of $h(x)$: $$ h(x)=1\cdot x(x−1)(x−2)(x−3)(x−4)(x−5) $$ $$ h(6)=1\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=720 $$ $$ f(6)=h(6) + (3\cdot6 +4)\to f(6)=720+22=742 $$

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Let f(x) be the ponomial P(x) - x^6. Then f(1) = 7 - 1^6 = 6 f(2) = 10 - 2^6 = -54 f(3) = 13 - 3^6 = -716 f(4) = 16 - 4^6 = -4080 f(5) = 19 - 5^6 = -15606

A 4th degree ponomial determined by these points is -140x^4+1050x^3-3101x^2+3993x-1796.

P(6) = f(6) + 6^6 = 2542.

If you chose some coefficient for x^5 you can get any value you wish for P(6).

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hi.. can you please explain how you got the 4th degree polynomial? –  Ankesh Aug 15 '13 at 16:55

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