Question

For some constants $a$ and $b,$ let \[f(x) = \left\{
\begin{array}{cl}
9 - 2x & \text{if } x \le 3, \\
ax + b & \text{if } x > 3.
\end{array}
\right.\]The function $f$ has the property that $f(f(x)) = x$ for all $x.$ What is $a + b?$

Answer 1

if f(f(x)) = x then f(x) is its own inverse.
Note that (3,3) is on the graph.
f^-1(x) = -1/2 x + 9/2 for x <= 3

These two functions are inverses also for x > 3
Looks like a+b = 4

Answer 2

I can't really say the build-up of it because it's not mine, but the answer is 7/3 :D Hope that helps

Answer 3

The answer is actually 4, Burrito you got it way wrong.

Answer 4

Setting $x = 0,$ we get $f(0) = 9.$ Since $9 > 3,$ $f(9) = 9a + b.$ Hence,$$f(f(0)) = f(9) = 9a + b.$$But $f(f(x)) = x$ for all $x,$ so $9a + b = 0.$

Setting $x = 1,$ we get $f(1) = 7.$ Since $7 > 3,$ $f(7) = 7a + b.$ Hence,$$f(f(1)) = f(7) = 7a + b.$$But $f(f(x)) = x$ for all $x,$ so $7a + b = 1.$

Subtracting the equations $9a + b = 0$ and $7a + b = 1,$ we get $2a = -1,$ so $a = -1/2.$ From $9a + b = 0,$ we get $b = -9a = 9/2.$ Hence,$$a + b = -1/2 + (9/2) = \boxed{4}.$$

Answer 5

the answer is 7/3.
Setting $x = 3,$ we get $f(3) = -1.$ Since $-1 < 2,$ $f(-1) = -a + b.$ Hence, $f(f(3)) = f(-1) = -a + b.$ But $f(f(x)) = x$ for all $x,$ so $-a + b = 3.$

Setting $x = 4,$ we get $f(4) = -4.$ Since $-4 < 2,$ $f(-4) = -4a + b.$ Hence, $f(f(4)) = f(-4) = -4a + b.$ But $f(f(x)) = x$ for all $x,$ so $-4a + b = 4.$

Subtracting the equations $-a + b = 3$ and $-4a + b = 4,$ we get $3a = -1,$ so $a = -1/3.$ From $-a + b = 3,$ we get $b = a + 3 = 8/3.$ Hence,$$a + b = (-1/3) + 8/3 = \boxed{7/3}.$$