November 7th, 2018, 05:38 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  What method can be used to work with embedded functions as shown in this problem?
The problem states the following: $\textrm{Given:}$ $$\phi\left( 2x+1\right)=6x10$$ $$\phi\left(f\left(x\right)3\right)=3x4$$ $\textrm{Find:}$ $f\left(\frac{1}{6}\right)$ The problem lies in the fact that the function $f\left(x\right)$ is embedded into $\phi$ (nested) and therefore it cannot be worked out in a straight manner. So I thought that the key to solve this would be to use inverse function so I could use then what is inner and solve a two equation system. By following that route I went this way: $\phi\left( 2x+1\right)=6x10$ $\phi^{1}\left(\phi\left( 2x+1\right)\right)=\phi^{1}\left(6x10\right)$ $2x+1= \frac{y+10}{6}$ $12x+610=y$ $y=12x4$ Then going with the second function which has embedded the $f\left(x\right)$: $\phi^{1}\left(\phi\left(f\left(x\right)3\right)\right)=\phi^{1}\left(3x4\right)$ $f\left(x\right)3=\frac{y+4}{3}$ $f\left(x\right)=\frac{y+4}{3}+3$ Then all what would be left to do is to replace the y into the above equation, isn't it? So I found: $f\left(x\right)=\frac{(12x4)+4}{3}+3=4x+3$ Then by inserting the value which is to be "evaluated" in the function would make it into: $f\left(\frac{1}{6})\right)=4\left(\frac{1}{6}\right)+3=\frac{2}{3}+3=\frac{7}{3}$ Therefore the answer would be $\frac{7}{3}$, however this does not appear within the alternatives in my book; to which given are: $37/6$, $35/4$, $35/6$, $37/4$, $35/6$. Could it be that Am I misinterpreting something or using the inverse function not properly?. I'd like someone could help me with this as I'm confused on which way should be recommended to solve these kinds of problems. 
November 7th, 2018, 07:54 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489 
Let's use substitution of variables. $\text {LET: } w = 2x + 1, \ u = \phi (w), \text { and } y = f(x)  3.$ $w = 2x + 1 \implies x = \dfrac{w  1}{2}.$ $u = \phi (w) = 6x  10 = 6 * \dfrac{w  1}{2}  10 = 3w  13.$ $\therefore \phi ( f(x)  3) = \phi ( f(y)) = 3y  13 = 3(f(x)  3)  13 = 3 * f(x)  22.$ $\text {And } \phi ( f(x)  3) = 3x  4 \implies 3 * f(x)  22 = 3x  4 \implies f(x) = x + 6.$ Let's check. $\phi (w) = 3w  13 \implies \phi (2x + 1) = 3(2x + 1)  13 = 6x + 3  13 = 6x  10.$ Looks good so far. $\phi (w) = 3w  13 \text { and } f(x) = x + 6 \implies \phi (f(x)  3) = \phi (x + 6  3) = \phi (x + 3) =$ $3(x + 3)  13 = 3x  4.$ Looks good as well. $f(x) = x + 6 \implies f \left (\ \dfrac{1}{6} \right ) = \ \dfrac{1}{6} + 6 = \dfrac{35}{36}.$ 
November 8th, 2018, 03:07 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,878 Thanks: 1835 
I'll assume that the given equations hold for all values of $x$. Putting $x = \frac16$ in the second equation gives $\phi\left(f\left(\frac16\right)  3\right) = 3\left(\frac16\right)  4 = \frac92$. To get a righthand side of $\frac92$ in the first equation, one must put $x = \frac{11}{12}$, which gives $\phi\left(\frac{17}{6}\right) = \frac92$. It follows that $f\left(\frac16\right)  3 = \frac{17}{6}$, so $f\left(\frac16\right) = \frac{17}{6} + 3 = \frac{35}{6}$. 
November 8th, 2018, 09:55 AM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489  Quote:
$\phi(x) = 3x  13 \text { and } f(x) = x + 6.$ $\therefore \phi(2x + 1) = 3(2x + 1)  13 = 6x + 3  13 = 6x  10.$ $\text {And } \phi ( f(x)  3) = \phi (x + 6  3) = \phi (x + 3) = 3(x + 3)  13 = 3x + 9  13 = 3x  4.$ $\therefore f \left ( \ \dfrac{1}{6} \right ) = \ \dfrac{1}{6} + 6 = \dfrac{35}{36}.$ That seems to me to satisfy the conditions of the problem, but it may not be the only way to satisfy the conditions of the problem. According to skipjack's method, $\phi \left ( f \left (  \dfrac{1}{6} \right )  3 \right ) = 3 * \left ( \ \dfrac{1}{6} \right )  4 = \ \dfrac{9}{2}.$ There can be no doubt the above satisfies the conditions of the problem. $6x  10 = \ \dfrac{9}{2} \iff x = \dfrac{11}{12}.$ No one can argue with that. $\therefore \phi (2x + 1) = 6x  10 \implies \phi \left ( 2 * \dfrac{11}{12} + 1 \right ) = 6 * \dfrac{11}{12}  10 \implies \phi \left ( \dfrac{17}{6} \right ) = \ \dfrac{9}{2}.$ Inarguable. There is now, as far as I can tell, a hidden assumption, namely that $\phi \left ( \dfrac{17}{6} \right ) = \ \dfrac{9}{2} = \phi \left ( f \left (\ \dfrac{1}{6} \right )  3 \right ) \implies \dfrac{17}{6} = f \left (  \ \dfrac{1}{6} \right )  3 \implies f \left ( \ \dfrac{1}{6} \right ) = \dfrac{35}{6}.$ But the first conclusion does not generally follow from the initial premise. $g(x) = x^2 \implies g(\ 2) = 4 = g(2) \not \implies \ 2 = 2.$ Of course, it is true for some functions that we can say that equality of the function entails equality of the arguments. For example, $log_a(b) = log_a(c) \iff b = c.$ So my objection may not be valid in this case because we have no clue what phi of x and f(x) are yet. However, when we worked with equation 1, we assumed x = 11/12. When we worked with equation 2, we assumed x =  1/6. That makes me dubious that we can equate anything. It appears that we are saying $a \ne b \text { and } g(a) = g(h(b)) \implies a = b.$ Skipjack knows 1000 times more math than I do so I am probably all wrong, but I am having a lot of trouble following his argument here.  
November 8th, 2018, 03:37 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,878 Thanks: 1835 
When determining that x must be 11/12 (and that this implies the lefthand side's argument is 17/6), one also discovers that $\phi$ is invertible (which justifies the next step). My use of the word "must" was intended to convey this.

November 8th, 2018, 03:40 PM  #6  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Quote:
$\phi(w)=3w13$ At this point I thought, Why not just replace in a straight manner $\phi(x)=3x13$ However this "$x$" variable was already in use and it could had been confused with was intended to find. Therefore you used "$y$" for $f(x)3$ The rest was just to evaluate this in the initial function and voilĂ the answer is obtained. At the beginning the check steps which you included were a bit confusing as I'm not very used to follow a set of three consecutive equal signs (this should not be understood as trying to say that they were unnecessary). But with some diligent effort I figured out that it checks with what was given. This riddle seems did not entitled much problem after all, since it was a set of two equations, however it could had been complicated if it happened more than two functions. Hopefully in my book I haven't found too complicated problems so far in this particular case.  
November 8th, 2018, 03:41 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489  Quote:
I guess I am still wondering about $g(x) = f(x)  3 \text { and } \phi \left ( \dfrac{11}{12} \right ) = \phi \left ( g \left (\ \dfrac{1}{6}\right ) \right ).$ This is not a case where we have the same invertible function on both sides of an equation. On one side we have an invertible function. On the other side we have a composition of that invertible function and another invertible function. In any case, I am beginning to think romsek was correct: there may not be a unique answer because my answer seems to work.  
November 8th, 2018, 03:49 PM  #8 
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489 
@ chemist. Skipjack seems to have found another answer so perhaps there is not a unique answer. That would take us back to romsek's post (since deleted) saying that there was insufficient information to find a unique answer. 
November 8th, 2018, 03:52 PM  #9  
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Quote:
 
November 8th, 2018, 03:59 PM  #10 
Member Joined: Jun 2017 From: Lima, Peru Posts: 52 Thanks: 1 Math Focus: Calculus  Sorry I wasn't aware of that. But from looking at the messages it looks that $x=\frac{11}{12}$. But does it means that there are other solutions as well?. I missed this part. Which would be the other solution?.


Tags 
embedded, functions, method, problem, shown, work 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Assorted questions, with no work shown.  LIPTON  Physics  1  June 28th, 2012 10:33 AM 
Verifying answer (with work shown)  Kimmysmiles0  Algebra  2  April 29th, 2012 09:38 PM 
Compound interest  no work shown  Arley  Algebra  1  March 22nd, 2012 08:51 PM 
Weird Embedded Cubes Problem  Recipe  Algebra  3  March 17th, 2010 07:10 PM 
Compound interest  no work shown  Arley  Calculus  0  December 31st, 1969 04:00 PM 