Find $a+b$ given $f(f(x)) = x$ where $f$ is piecewise defined

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 $a + b?$

At first, I thought of plugging $x=1,$ to get $a+b,$ but that wouldn't be much of a help, as $1<3.$ Solution?

Answer 1

Assuming $a$ and $b$ exist such that they satisfy the given conditions. For $x = 0$, we get $f(x) = 9$.

It is given that, $f(f(x)) = x$

$$\therefore 9a + b = 0$$

Simillarly using $x = 1$, and $f(f(x)) = x$ we get,

$$7a + b = 1$$

Solving these two equations gives:

$$a = \frac{-1}{2}, b = \frac{9}{2}$$

Now let's show that with these values for $a$ and $b$ $f(f(x)) = x$, $\forall x \in \mathbb{R}$.

Case I: $x < 3$

It is obvious that $x < 3 \implies f(x) > 3$.

Therefore, $f(f(x)) = f(9-2x) = \frac{-1}{2}(9-2x) + \frac{9}{2} = x$

Case II: $x = 3$

$x = 3 \implies f(x) = 3 \implies f(f(x)) = 3 = x$

Case II: $x > 3$

Again it is obvious that $x > 3 \implies f(x) < 3$.

Therefore, $f(f(x)) = f(ax + b) = 9 - 2 (\frac{-1}{2}x + \frac{9}{2}) = x$

Answer 2

You need to compute two values that are less than $3$. Why? because this is what will output a number - and you will know if it is greater or less than $3$.

Let's choose $x=1$ and $x=2$ for example:

$$f(f(1)) = f(7) = 7a+b = 1$$

$$f(f(2)) = f(5) = 5a+b = 2$$

Now we have a system of equations (2 eq. and 2 unknowns)
This should be straight forward from here:

$$ a = - \frac{1}{2}$$ $$ b = \frac{9}{2}$$

And thus: $$a + b = \frac{8}{2} = 4$$

Answer 3

There is yet another simple solution $$f(x) = \left\{ \begin{array}{cl} 9 - 2x & \text{if } x \le 3 \\ ax + b & \text{if } x > 3. \end{array} \right.$$ $$f^{-1}(x) = \left\{ \begin{array}{cl} \frac{9 - x}{2} & \text{if } x \ge 3 \\ \frac{x-b}{a} & \text{if } x < 3. \end{array} \right.$$

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$f(f(x))=x \implies f(x)=f^{-1}(x)$

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Now $$\boxed{\frac{9 - x}{2}=ax+b},x>3$$ or $$\boxed{\frac{x-b}{a}=9-2x},x<3$$

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Now get two equations and with any 2 values of $x$ in respective domains to obtain 2 equations giving $a,b$

Example

Take $x=4$ $$4a+b=\frac{5}{2}$$ Take $x=1$ $$7a+b=1$$

Solve to get $\boxed{a=\frac{-1}{2}}$ and $\boxed{b=\frac{9}{2}}$

Answer 4

As $(0,9)$ and $(4.5,0)$ are points of $9-2x$ we notice that $(9,0)$ and $(0,4.5)$ are two points on $ax+b$. From here $ax+b=-0.5x+4.5$.