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Composing two of three given functions to obtain $2^x+1 - 1$

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0

$begingroup$

A problem I'm facing involves three given functions. I need to compose two of them to obtain $2^x+1-1$. The three functions I'm allowed to use are

• $f(x) = 2x-1$


• $g(x) = 1/x$


• $h(x) = 2^x$

It would be great if there was a detailed explanation as to how you got the answer too. Thanks!

algebra-precalculus functions function-and-relation-composition

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

V11V11

196

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it $$2^x+1-1$$?
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I couldn't format it well.
  $endgroup$
  – V11
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are three ways to try composing two functions, and 6 ways to try composing all three. If those failed you could try further compositions I suppose, but it's pretty obvious any further compositions will fail. You only need to compose two functions, so just try the three combinations. This is not a homework-solving site, so you need to show some effort and context for your problem.
  $endgroup$
  – Brevan Ellefsen
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I was thinking maybe fh(x+1). Is that right? I'm not very sure.
  $endgroup$
  – V11
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For future reference, here's a MathJax reference - the markup language used to render math text here. math.meta.stackexchange.com/questions/5020/…
  $endgroup$
  – Eevee Trainer
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

A problem I'm facing involves three given functions. I need to compose two of them to obtain $2^x+1-1$. The three functions I'm allowed to use are

• $f(x) = 2x-1$


• $g(x) = 1/x$


• $h(x) = 2^x$

It would be great if there was a detailed explanation as to how you got the answer too. Thanks!

algebra-precalculus functions function-and-relation-composition

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

V11V11

196

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it $$2^x+1-1$$?
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I couldn't format it well.
  $endgroup$
  – V11
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There are three ways to try composing two functions, and 6 ways to try composing all three. If those failed you could try further compositions I suppose, but it's pretty obvious any further compositions will fail. You only need to compose two functions, so just try the three combinations. This is not a homework-solving site, so you need to show some effort and context for your problem.
  $endgroup$
  – Brevan Ellefsen
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I was thinking maybe fh(x+1). Is that right? I'm not very sure.
  $endgroup$
  – V11
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For future reference, here's a MathJax reference - the markup language used to render math text here. math.meta.stackexchange.com/questions/5020/…
  $endgroup$
  – Eevee Trainer
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

A problem I'm facing involves three given functions. I need to compose two of them to obtain $2^x+1-1$. The three functions I'm allowed to use are

• $f(x) = 2x-1$


• $g(x) = 1/x$


• $h(x) = 2^x$

It would be great if there was a detailed explanation as to how you got the answer too. Thanks!

algebra-precalculus functions function-and-relation-composition

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

V11V11

196

$endgroup$

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

V11V11

196

share|cite|improve this question

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

V11V11

196

edited 2 days ago

Eevee Trainer

7,80721339

edited 2 days ago

Eevee Trainer

7,80721339

asked 2 days ago

V11V11

196

asked 2 days ago

V11V11

196

3 Answers
3

active

oldest

votes

3

$begingroup$

Hint:

Consider the functions $f$ and $h$. This answer will be more complicated than simply plugging things in, you'll need to make use of some algebraic properties.

In particular, the property that $$a^b a^c = a^b+c$$

---

Solution:

So, $f(x) = 2x-1, h(x) = 2^x$. We know we want to get two to the something, so we should apply $h$ at some point, and we kind of want a $-1$ to appear at some point, so that's why we choose $f$.

Sadly beyond that stuff like this is largely guess and check (such as other possible functions or the order to apply them); I'll say right off that the right composition is $f circ h$.

This gives us the function composition

$$(fcirc h)(x) = fleft( h(x) right) = 2cdot 2^x - 1$$

We make use of the property I noted earlier: $2cdot 2^x = 2^x+1$. Thus

$$(fcirc h)(x) = 2^x+1 - 1$$

share|cite|improve this answer

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

$endgroup$

add a comment | 

0

$begingroup$

Here's the detailed explanation as to how to get the answer:

Compute all applicable compositions, i.e., $f(g(x))$, $g(f(x))$, $f(h(x))$, $h(f(x))$, $g(h(x))$, $h(g(x))$. Then check if any of these equals (or can be simplified to) the desired function of $x$.

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We must not compute all these possibilites, since the solution should be $$2^x+1-1$$
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner Of course, we can stop this extremely finite task as soon as we happen to run into a valid solution. -- Of course one could make one's life a lot easier by observing that the desired composition has $0$ in ist range and is defined at $0$, hence $g$ cannot occur in the composition. Moreover, the desired function has negative numbers in its range, which excludes $h$ as outer function. Hence $fcirc h$ is the only candidate, and if we rely on the validity of the problem statement, we can write it down without further testing ...
  $endgroup$
  – Hagen von Eitzen
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

But $$h(f(x))=2^2x-1,h(g(x))=2^1/x$$ which gives not your solution, so $$f(h(x))$$ leads to your solutions

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866

$endgroup$

add a comment | 

Your Answer

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3

$begingroup$

Hint:

Consider the functions $f$ and $h$. This answer will be more complicated than simply plugging things in, you'll need to make use of some algebraic properties.

In particular, the property that $$a^b a^c = a^b+c$$

---

Solution:

So, $f(x) = 2x-1, h(x) = 2^x$. We know we want to get two to the something, so we should apply $h$ at some point, and we kind of want a $-1$ to appear at some point, so that's why we choose $f$.

Sadly beyond that stuff like this is largely guess and check (such as other possible functions or the order to apply them); I'll say right off that the right composition is $f circ h$.

This gives us the function composition

$$(fcirc h)(x) = fleft( h(x) right) = 2cdot 2^x - 1$$

We make use of the property I noted earlier: $2cdot 2^x = 2^x+1$. Thus

$$(fcirc h)(x) = 2^x+1 - 1$$

share|cite|improve this answer

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

$endgroup$

add a comment | 

3

$begingroup$

Hint:

Consider the functions $f$ and $h$. This answer will be more complicated than simply plugging things in, you'll need to make use of some algebraic properties.

In particular, the property that $$a^b a^c = a^b+c$$

---

Solution:

So, $f(x) = 2x-1, h(x) = 2^x$. We know we want to get two to the something, so we should apply $h$ at some point, and we kind of want a $-1$ to appear at some point, so that's why we choose $f$.

Sadly beyond that stuff like this is largely guess and check (such as other possible functions or the order to apply them); I'll say right off that the right composition is $f circ h$.

This gives us the function composition

$$(fcirc h)(x) = fleft( h(x) right) = 2cdot 2^x - 1$$

We make use of the property I noted earlier: $2cdot 2^x = 2^x+1$. Thus

$$(fcirc h)(x) = 2^x+1 - 1$$

share|cite|improve this answer

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

$endgroup$

add a comment | 

$begingroup$

Hint:

Consider the functions $f$ and $h$. This answer will be more complicated than simply plugging things in, you'll need to make use of some algebraic properties.

In particular, the property that $$a^b a^c = a^b+c$$

---

Solution:

So, $f(x) = 2x-1, h(x) = 2^x$. We know we want to get two to the something, so we should apply $h$ at some point, and we kind of want a $-1$ to appear at some point, so that's why we choose $f$.

Sadly beyond that stuff like this is largely guess and check (such as other possible functions or the order to apply them); I'll say right off that the right composition is $f circ h$.

This gives us the function composition

$$(fcirc h)(x) = fleft( h(x) right) = 2cdot 2^x - 1$$

We make use of the property I noted earlier: $2cdot 2^x = 2^x+1$. Thus

$$(fcirc h)(x) = 2^x+1 - 1$$

share|cite|improve this answer

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

$endgroup$

share|cite|improve this answer

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

answered 2 days ago

Eevee TrainerEevee Trainer

7,80721339

0

$begingroup$

Here's the detailed explanation as to how to get the answer:

Compute all applicable compositions, i.e., $f(g(x))$, $g(f(x))$, $f(h(x))$, $h(f(x))$, $g(h(x))$, $h(g(x))$. Then check if any of these equals (or can be simplified to) the desired function of $x$.

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We must not compute all these possibilites, since the solution should be $$2^x+1-1$$
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner Of course, we can stop this extremely finite task as soon as we happen to run into a valid solution. -- Of course one could make one's life a lot easier by observing that the desired composition has $0$ in ist range and is defined at $0$, hence $g$ cannot occur in the composition. Moreover, the desired function has negative numbers in its range, which excludes $h$ as outer function. Hence $fcirc h$ is the only candidate, and if we rely on the validity of the problem statement, we can write it down without further testing ...
  $endgroup$
  – Hagen von Eitzen
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Here's the detailed explanation as to how to get the answer:

Compute all applicable compositions, i.e., $f(g(x))$, $g(f(x))$, $f(h(x))$, $h(f(x))$, $g(h(x))$, $h(g(x))$. Then check if any of these equals (or can be simplified to) the desired function of $x$.

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We must not compute all these possibilites, since the solution should be $$2^x+1-1$$
  $endgroup$
  – Dr. Sonnhard Graubner
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dr.SonnhardGraubner Of course, we can stop this extremely finite task as soon as we happen to run into a valid solution. -- Of course one could make one's life a lot easier by observing that the desired composition has $0$ in ist range and is defined at $0$, hence $g$ cannot occur in the composition. Moreover, the desired function has negative numbers in its range, which excludes $h$ as outer function. Hence $fcirc h$ is the only candidate, and if we rely on the validity of the problem statement, we can write it down without further testing ...
  $endgroup$
  – Hagen von Eitzen
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Here's the detailed explanation as to how to get the answer:

Compute all applicable compositions, i.e., $f(g(x))$, $g(f(x))$, $f(h(x))$, $h(f(x))$, $g(h(x))$, $h(g(x))$. Then check if any of these equals (or can be simplified to) the desired function of $x$.

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

$endgroup$

share|cite|improve this answer

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

answered 2 days ago

Hagen von EitzenHagen von Eitzen

282k23272507

0

$begingroup$

But $$h(f(x))=2^2x-1,h(g(x))=2^1/x$$ which gives not your solution, so $$f(h(x))$$ leads to your solutions

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866

$endgroup$

add a comment | 

0

$begingroup$

But $$h(f(x))=2^2x-1,h(g(x))=2^1/x$$ which gives not your solution, so $$f(h(x))$$ leads to your solutions

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866

$endgroup$

add a comment | 

$begingroup$

But $$h(f(x))=2^2x-1,h(g(x))=2^1/x$$ which gives not your solution, so $$f(h(x))$$ leads to your solutions

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866

answered 2 days ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.7k42866