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November 7th, 2018, 05:38 PM   #1
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Question What method can be used to work with embedded functions as shown in this problem?

The problem states the following:

$\textrm{Given:}$

$$\phi\left( 2x+1\right)=6x-10$$
$$\phi\left(f\left(x\right)-3\right)=3x-4$$

$\textrm{Find:}$
$f\left(-\frac{1}{6}\right)$

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:

$\phi\left( 2x+1\right)=6x-10$

$\phi^{-1}\left(\phi\left( 2x+1\right)\right)=\phi^{-1}\left(6x-10\right)$

$2x+1= \frac{y+10}{6}$

$12x+6-10=y$

$y=12x-4$

Then going with the second function which has embedded the $f\left(x\right)$:

$\phi^{-1}\left(\phi\left(f\left(x\right)-3\right)\right)=\phi^{-1}\left(3x-4\right)$

$f\left(x\right)-3=\frac{y+4}{3}$

$f\left(x\right)=\frac{y+4}{3}+3$

Then all what would be left to do is to replace the y into the above equation, isn't it?

So I found:

$f\left(x\right)=\frac{(12x-4)+4}{3}+3=4x+3$

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.
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November 7th, 2018, 07:54 PM   #2
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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.

$f(x) = x + 6 \implies f \left (-\ \dfrac{1}{6} \right ) = -\ \dfrac{1}{6} + 6 = \dfrac{35}{36}.$
Thanks from topsquark and Chemist116
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November 8th, 2018, 03:07 AM   #3
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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 right-hand 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 $f\left(-\frac16\right) = \frac{17}{6} + 3 = \frac{35}{6}$.
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November 8th, 2018, 09:55 AM   #4
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Originally Posted by skipjack View Post
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 right-hand 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 $f\left(-\frac16\right) = \frac{17}{6} + 3 = \frac{35}{6}$.
According to my method

$\phi(x) = 3x - 13 \text { and } f(x) = x + 6.$

$\therefore \phi(2x + 1) = 3(2x + 1) - 13 = 6x + 3 - 13 = 6x - 10.$

$\text {And } \phi ( f(x) - 3) = \phi (x + 6 - 3) = \phi (x + 3) = 3(x + 3) - 13 = 3x + 9 - 13 = 3x - 4.$

$\therefore f \left ( -\ \dfrac{1}{6} \right ) = -\ \dfrac{1}{6} + 6 = \dfrac{35}{36}.$

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.
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November 8th, 2018, 03:37 PM   #5
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When determining that x must be 11/12 (and that this implies the left-hand 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.
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November 8th, 2018, 03:40 PM   #6
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Quote:
Originally Posted by JeffM1 View Post
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.

$f(x) = x + 6 \implies f \left (-\ \dfrac{1}{6} \right ) = -\ \dfrac{1}{6} + 6 = \dfrac{35}{36}.$
I'd have to say initially I felt dumbfounded as I couldn't understand what was going on by looking at different variables and subsequent steps. But by the fourth time I looked into it it kind of made sense by looking at this equation:

$\phi(w)=3w-13$

At this point I thought, Why not just replace in a straight manner

$\phi(x)=3x-13$

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.
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November 8th, 2018, 03:41 PM   #7
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Quote:
Originally Posted by skipjack View Post
When determining that x must be 11/12 in my solution, one also discovers that $\phi$ is invertible (which justifies the next step). My use of the word "must" was intended to convey this.
Yes, my solution for $\phi(x)$ is invertible so I am with you there. Furthermore, my solution for f(x) is also invertible.

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.
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November 8th, 2018, 03:49 PM   #8
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@ 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.
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November 8th, 2018, 03:52 PM   #9
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Quote:
Originally Posted by JeffM1 View Post
According to my method

$\phi(x) = 3x - 13 \text { and } f(x) = x + 6.$

$\therefore \phi(2x + 1) = 3(2x + 1) - 13 = 6x + 3 - 13 = 6x - 10.$

$\text {And } \phi ( f(x) - 3) = \phi (x + 6 - 3) = \phi (x + 3) = 3(x + 3) - 13 = 3x + 9 - 13 = 3x - 4.$

$\therefore f \left ( -\ \dfrac{1}{6} \right ) = -\ \dfrac{1}{6} + 6 = \dfrac{35}{36}.$

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.
I don't how if it adds to the answer this article about functions which does mentions about injective, surjective and bijective ones. Which is explained in details here. Then for all of this to happen a function must be bijective?. Initially what I asked was if this problem could had been solved using "inverse functions" but it looks that this is what it has been done but just I didn't figured out?. Overall I think any method either by evaluating $-\frac{1}{6}$ at the beginning and developing the solution from there or let it at the end, both methods work and that suffices my needs, although as mentioned I was hoping somebody could explain me if using inverse function would had "shortened" the steps.
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November 8th, 2018, 03:59 PM   #10
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Originally Posted by JeffM1 View Post
@ 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.
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?.
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