2
$\begingroup$

How can one calculate $f(3)$ when $f(f(x))=x^2-5x+9$

I tried this: $f(f(3))=3$

I'm stuck here.

$\endgroup$
2
  • $\begingroup$ Setting $x^2-5x+9=t$ then we get $x_{1,2}=5/2\pm \sqrt {t-11/4}$ and you will get $f(t)=...$ $\endgroup$ Dec 7, 2015 at 20:15
  • 2
    $\begingroup$ Your initial statement should be phrased as "Calculate $f(3)$ where $f(f(x)) = x^2 - 5x + 9$." ${}\qquad{}$ $\endgroup$ Dec 7, 2015 at 20:18

3 Answers 3

6
$\begingroup$

$f(f(f(x))) = f(x^2 - 5x +9) = f(x)^2 - 5 f(x) + 9$.

set $x = 3$, then $f(3) = f(3)^2 - 5 f(3) + 9$. then $f(3) = 3$

$\endgroup$
2
  • 1
    $\begingroup$ how did you get $f(x)^2 - 5 f(x) + 9$.? $\endgroup$
    – user233658
    Dec 7, 2015 at 20:19
  • 1
    $\begingroup$ using $f(f(y)) = y^2 - 5y +9$, replace $y$ as $f(x)$. @user233658 $\endgroup$
    – Yimin
    Dec 7, 2015 at 20:20
4
$\begingroup$

Since $f(f(3))=3^2-5\cdot 3+9=3$, we have $$f(3)=f(f(f(3)))=(f(3))^2-5f(3)+9,$$ i.e. $$(f(3)-3)^2=0.$$ So, $f(3)=3$.

$\endgroup$
4
$\begingroup$

If $f(3) = y$, then $f(y) = f(f(3)) = 3$, and $f(f(y)) = f(3) = y$.
But $0 = f(f(y)) - y = y^2 - 6 y + 9 = (y - 3)^2$, so $y = 3$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .