Calculate $f(3)$ such that $f(f(x))=x^2-5x+9$

Question

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.

Setting $x^2-5x+9=t$ then we get $x_{1,2}=5/2\pm \sqrt {t-11/4}$ and you will get $f(t)=...$ — Dr. Sonnhard Graubner, Dec 07 '15 at 20:15
Your initial statement should be phrased as "Calculate $f(3)$ where $f(f(x)) = x^2 - 5x + 9$." ${}\qquad{}$ — Michael Hardy, Dec 07 '15 at 20:18

score 6 · Accepted Answer · answered Dec 07 '15 at 20:16

6

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

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

answered Dec 07 '15 at 20:16

Yimin

1

how did you get $f(x)^2 - 5 f(x) + 9$.? – user233658 Dec 07 '15 at 20:19
1

using $f(f(y)) = y^2 - 5y +9$, replace $y$ as $f(x)$. @user233658 – Yimin Dec 07 '15 at 20:20

score 4 · Answer 2 · answered Dec 07 '15 at 20:16

4

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, $f(3)=3$.

answered Dec 07 '15 at 20:16

mathlove

score 4 · Answer 3 · answered Dec 07 '15 at 20:17

4

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 $y = 3$.

answered Dec 07 '15 at 20:17

Robert Israel

3 Answers3