# Math Help - Functions

1. ## Functions

1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.

First time seeing this. Somehow tell me how to approach it, I don't need the answer.

2. 2.
Since we are given $\frac{-d}{c}$ isn't in the domain, we know that any number of the form $\frac{-ad}{c}+b$ isn't in the range of f.

3. For the first one, try substituting a in for x, and then 1/a in for x. Solve the resulting system of equations for f(a) or f(1/a) and see what pops out.

4. ## Rational function.

Originally Posted by awkwardnerd
1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.

First time seeing this. Somehow tell me how to approach it, I don't need the answer.
#2.
$\displaystyle f(f(x))=x\quad\implies\quad f^{-1}(x)=f(x)$.

$\displaystyle f(x)$ has a vertical asymptote at $\displaystyle x=-\,{{d}\over{c}}$ and a horizontal asymptote at $\displaystyle f(x)={{a}\over{c}}$. Therefore, $\displaystyle f^{-1}(x)$ has a horizontal asymptote at $\displaystyle x=-\,{{d}\over{c}}$ and a vertical asymptote at $\displaystyle f^{-1}(x)={{a}\over{c}}$.

Thus $\displaystyle f^{-1}(x)=f(x)\quad\implies\quad -\,{{d}\over{c}}={{a}\over{c}}$ so that $\displaystyle a=-d$.

Also, the zero of $\displaystyle f(x)$ is equal to the y-intercept of $\displaystyle f^{-1}(x)$, and vice-versa, but I see that this just gives $\displaystyle d=-a$ again.

So, we have that $\displaystyle f(x)={{ax+b}\over{cx-a}}$.

We also know that $\displaystyle f(x)=x$ for $\displaystyle x=19,\ 97\,.$ Solve $\displaystyle f(x)=x$ and that should give us $\displaystyle x=58\pm39=97,\ 19\,.$

$\displaystyle {{ax+b}\over{cx-a}}=x$

$\displaystyle ax+b=x(cx-a)$

$\displaystyle cx^2-2ax-b=0$

$\displaystyle x={{2a\pm\sqrt{4a^2+4bc}}\over{2c}}={{a\pm\sqrt{a^ 2+bc}}\over{c}} ={{a}\over{c}}\pm{{\sqrt{a^2+bc}}\over{c}}=58\pm39$

Thus, $\displaystyle a=58c$ and $\displaystyle {{\sqrt{a^2+bc}}\over{c}}=39\ \ \implies\ \ b=(39^2-58^2)c \ \implies\ \ b=-(19)(97)c$

Therefore, $\displaystyle f(x)={{58cx-(19)(97)c}\over{cx-58c}}.$ Cancel $\displaystyle c$.

$\displaystyle f(x)={{58x-1843}\over{x-58}}.$

5. Originally Posted by DrSteve
For the first one, try substituting a in for x, and then 1/a in for x. Solve the resulting system of equations for f(a) or f(1/a) and see what pops out.
Plugging in $a$ for x: $f(a)+2f(\frac{1}{a})=3a$
Pulgging in $\frac{1}{a}$ for x: $f(\frac{1}{a})+2f(a)=\frac{3}{a}$

Solving both of these for $f(a)$:
$f(a)=3a-2f(\frac{1}{a})=\frac{3}{2a}-\frac{f(\frac{1}{a})}{2}$
Also, $f(a)=f(-a) \quad\implies\quad -3a-2f(\frac{-1}{a})=\frac{-3}{2a}-\frac{f(\frac{-1}{a})}{2}$
And $-3a-2f(\frac{1}{a})=\frac{-3}{2a}-\frac{f(\frac{1}{a})}{2}$

When I look at these all I get is $a=0$, which is excluded from the domain. I get the same thing when I solve for $f(\frac{1}{a})$. What am I missing?

6. Originally Posted by awkwardnerd
1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

First time seeing this. Somehow tell me how to approach it, I don't need the answer.

#1.
pflo's post gave me the idea for the solution.

$\displaystyle f(x)+2f\left({{1}\over{x}}\right)=3x\ \ \to\ \ -f(x)-2f\left({{1}\over{x}}\right)=-3x$

$\displaystyle f\left({{1}\over{x}}\right)+2f(x)={{3}\over{x}}\ \ \ \to\ \ \ 4f(x)+2f\left({{1}\over{x}}\right)=\ {{6}\over{x}}$

Add the equations on the right to eliminate $f\left({{1}\over{x}}\right),$ then solve for $\displaystyle f(x)$.

As it turns out, f is an odd function, so $\displaystyle f(a)=f(-a)$ implies that $\displaystyle f(a)=0$.

There are two values of a which satisfy this condition.

7. Originally Posted by SammyS
As it turns out, f is an odd function, so $\displaystyle f(a)=f(-a)$ implies that $\displaystyle f(a)=0$.
I agree that $f$ is an odd function because $f(-a)=f(a)$, but why would this imply that $f(a)=0$?

8. Originally Posted by pflo
I agree that $f$ is an odd function because $f(-a)=f(a)$, but why would this imply that $f(a)=0$?
No, the definition of odd function is that f(-a)= -f(a). If you know that f is an odd function and that f(-a)= f(a) then f(a)= -f(a) so that f(a)= 0.

9. Duh. Thanks. I wasn't thinking straight. If $f(-x)=f(x)$, then $f$ is an even function! But this isn't always true for this function: we're just looking for values of x where this would be true.

So this brings me to another question: how do we know that the function at issue here is an odd function?

10. Note that $f$ being odd always implies that $f(0) = 0$ (you don't need the other condition). Indeed, if $f(-x)=-f(x)$ for all $x$, then substituting a $0$ in for $x$ yields $f(-0)=-f(0)$. But $-0 = 0$ whence $f(0) = -f(0)$. Adding $f(0)$ to both sided yields $2f(0) = 0$. Finally, divide both sides by $2$ to get $f(0) = 0$.

Edit: I should clarify that $f$ being odd always implies that $f(0)=0$ whenever $0$ is in the domain of $f$.

11. Here is how we know that f is odd:

Adding the two equations in Sammy's post shows that $3f(x)=-3f(x)+\frac{6}{x}$, so that $f(x)=-f(x)+\frac{2}{x}$. Thus, $2f(x) = \frac{2}{x}$ , and $f(x) = \frac{1}{x}$ which is odd.