Let $f:\mathbb R \to \mathbb R$ be a function, such as : $3f(x+1)-2f(2-x) = x^2 + 14x - 5$. Find $f(x)$.

Question

Let $f:\mathbb R \to \mathbb R$ be a function, such as :

$$3f(x+1)-2f(2-x) = x^2 + 14x - 5$$

Find f(x)

Attempt :

Let $x+1 := y \Leftrightarrow x = y-1$. Then, we get :

$$3f(y) - 2f(3-y) = (y-1)^2 + 14(y-1) - 5$$

so I got $f(y)$ in the expression but still have to simplify it further.

I also observed that the two arguments of the functions $f$ become equal at :

$$y = 3-y \Leftrightarrow y = \frac{1}{2}$$

It should be an easy one since I found it across an Elementary Leaflet but my head has been stuck for now.

Any help would be appreciated !

Answer 1

Let $z=1-x$ & we have \begin{eqnarray*} 3f(2-z)-2f(1+z)=z^2-16z+10. \end{eqnarray*} Now let $z=x$ in the equation above & multiply this by $3$ and add $2$ times the first equation and we have \begin{eqnarray*} 5f(2-x)=5x^2-20x+20. \end{eqnarray*} So $\color{blue}{f(x)=x^2}$.

Answer 2

Note that $x+1=2-x$ if $x = 1/2$. Thus for $x = 1/2$ the equation says $f(3/2) = 9/4$. You can assign arbitrary values to $f(x)$ for $x > 3/2$, while for $x < 3/2$, $$f \left( x \right) =\frac13\,{x}^{2}+4\,x-6+\frac23\,f \left( 3-x \right) $$

Answer 3

Hint The substitution $x = 1-y$ gives

$$3 f(2-y) - 2 f(1+y) = (1-y)^2+ 14(1-y) - 5$$

Answer 4

I'd just assume an ansatz: $f(x)=ax^2+bx+c$.

This gives

$$ax^2 + (14a + 5b)x+(-5a-b+c) = x^2 +14x - 5.$$

Clearly, $a=1$ from the quadratic term, and then $b=0$ from the linear term. Then, $c=0$ from the constant term.

So it looks like $f(x)=x^2.$

Answer 5

We can make the two $f(a)$s equal by setting $x=1/2$

$$3f(x+1) -2f(2-x) = x^2 + 14x - 5$$ $$3f(3/2) - 2f(3/2) = (1/2)^2 + 14 \cdot 1/2-5$$ $$f(3/2) = 9/4$$

From which we can guess that $f(x) = x^2$