# Thread: need help solving this function....

1. ## need help solving this function....

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

2. ## Re: need help solving this function....

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

3. ## Re: need help solving this function....

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

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