A polynomial f(x) satisfies the equation f(x)+(x+1)3=2f(x+1). Find f(10).

Jan 27th 2013, 11:10 AM

emakarov

Re: need help solving this function....

Write $\displaystyle f(x)$ as $\displaystyle a_3x^3+a_2x^2+a_1x+a_0$ (you may also add $\displaystyle a_5x^5+a_4x^4$ or higher degrees), write the equation $\displaystyle f(x)+(x+1)^3=2f(x+1)$ in terms of $\displaystyle a_0$, ..., $\displaystyle a_3$ and equate the coefficients of the corresponding monomials. You'll get a system of four linear equations in $\displaystyle a_0$, ..., $\displaystyle a_3$.

Jan 27th 2013, 07:58 PM

earthboy

Re: need help solving this function....

ok, emakarov has the idea...
try, but if you want to learn the full technique here it is (Hi)

Spoiler:

your question: find $\displaystyle f(10)$ if
$\displaystyle 2f(x+1)-f(x)=3(x+1)$
now, you have stated that $\displaystyle f(x)$ is a polynomial, I am assuming it is a quadratic polynomial (this dosen't always work though...but school problems are generally that way(Wink) )
so let $\displaystyle f(x)=ax^2+bx+c$
so $\displaystyle f(x+1)=a(x+1)^2+b(x+1)+c$
now according to your equation,
$\displaystyle 2(a(x+1)^2+b(x+1)+c)-ax^2-bx-c=3x+3$
$\displaystyle \implies ax^2+4ax+2a+bx+2b+c=3x+3 $(you will get this by expanding and simplifying the above equation)
now you can see that the right hand side has no $\displaystyle x^2$ in it, so a must be equal to 0 in order to equate it.
if $\displaystyle a=0$, the equation shortens to $\displaystyle bx+2b+c=3x+3$
we know the equation to be true for all values of $\displaystyle x $
so put $\displaystyle x=0$, it becomes $\displaystyle 2b+c=3$
put $\displaystyle x=-1$, it becomes $\displaystyle b+c=0 $or $\displaystyle b=-c$ and so $\displaystyle 2b+c=-2c+c=-c=3$ or $\displaystyle c=-3$
so we get $\displaystyle c=-3,b=3,a=0$ and our function $\displaystyle f(x)$ becomes $\displaystyle 3x-3$.if $\displaystyle f(x)=3x-3$, $\displaystyle \boxed{f(10)=3*10-3=27} $