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Difference between revisions of "1980 AHSME Problems/Problem 14"
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As <math>f(x)=cx/2x+3</math>, we can plug that into <math>f(f(x))</math> and simplify to get <math>c^2x/2cx+6x+9 = x</math>
 
As <math>f(x)=cx/2x+3</math>, we can plug that into <math>f(f(x))</math> and simplify to get <math>c^2x/2cx+6x+9 = x</math>
 
. However, we have a restriction on x such that if <math>x=-3/2</math> we have an undefined function. We can use this to our advantage. Plugging that value for x into <math>c^2x/2cx+6x+9 = x</math> yields <math>c/2 = -3/2</math>, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that <math>c=-3</math>, which is answer choice <math>\boxed{A}</math>.
 
. However, we have a restriction on x such that if <math>x=-3/2</math> we have an undefined function. We can use this to our advantage. Plugging that value for x into <math>c^2x/2cx+6x+9 = x</math> yields <math>c/2 = -3/2</math>, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that <math>c=-3</math>, which is answer choice <math>\boxed{A}</math>.
 +
  
 
Alternatively, after simplifying the function to <math>c^2x/2cx+6x+9 = x</math>, multiply both sides by <math>2cx+6x+9</math> and divide by <math>x</math> to yield <math>c^2=2cx+6x+9</math>. This can be factored to <math>x(2c+6) + (3+c)(3-c) = 0</math>. This means that both <math>2c+6</math> and either one of <math>3+c</math> or <math>3-c</math> are equal to 0. <math>2c+6=0</math> yields <math>c=-3</math> and the other two yield <math>c=3,-3</math>. The clear solution is <math>c=-3</math>.
 
Alternatively, after simplifying the function to <math>c^2x/2cx+6x+9 = x</math>, multiply both sides by <math>2cx+6x+9</math> and divide by <math>x</math> to yield <math>c^2=2cx+6x+9</math>. This can be factored to <math>x(2c+6) + (3+c)(3-c) = 0</math>. This means that both <math>2c+6</math> and either one of <math>3+c</math> or <math>3-c</math> are equal to 0. <math>2c+6=0</math> yields <math>c=-3</math> and the other two yield <math>c=3,-3</math>. The clear solution is <math>c=-3</math>.

Revision as of 20:29, 24 April 2016

Problem

If the function $f$ is defined by \[f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} ,\] satisfies $x=f(f(x))$ for all real numbers $x$ except $-\frac{3}{2}$, then $c$ is

$\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$


Solution

As $f(x)=cx/2x+3$, we can plug that into $f(f(x))$ and simplify to get $c^2x/2cx+6x+9 = x$ . However, we have a restriction on x such that if $x=-3/2$ we have an undefined function. We can use this to our advantage. Plugging that value for x into $c^2x/2cx+6x+9 = x$ yields $c/2 = -3/2$, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that $c=-3$, which is answer choice $\boxed{A}$.


Alternatively, after simplifying the function to $c^2x/2cx+6x+9 = x$, multiply both sides by $2cx+6x+9$ and divide by $x$ to yield $c^2=2cx+6x+9$. This can be factored to $x(2c+6) + (3+c)(3-c) = 0$. This means that both $2c+6$ and either one of $3+c$ or $3-c$ are equal to 0. $2c+6=0$ yields $c=-3$ and the other two yield $c=3,-3$. The clear solution is $c=-3$.


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