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What is the range of this combined function?

Finding the Domain and Range of a function composition$x = sec 2y$, Find $dfrac dydx$ in terms of $x$. What about $pm$?Domain and range of an inverse functionWhy does the domain and range of $sqrt x$ contain only positive real numbers?What might this function be?when to use restrictions (domain and range) on trig functionsFinding the Range and the Domain of $f(x)=frac x^21-x$Finding the domain of $(f circ g)(x)$Confusion About Domain and Range of Linear Composite FunctionsRange of a function, with contradictory restriction

5

1

$begingroup$

I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.

Given $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, we are asked to find the domain and range of the combined function
$$(f circ g)(x)$$

My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:

Since $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.

Combining these two restrictions, my solution for the range is

$$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$

The given solution, however, is:

$$y in mathbbR mid y neq > 0 $$

I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?

algebra-precalculus functions

share|cite|improve this question

edited Apr 28 at 23:38

Calculemus

asked Apr 28 at 23:20

CalculemusCalculemus

439317

$endgroup$

add a comment | 

5

1

$begingroup$

I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.

Given $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, we are asked to find the domain and range of the combined function
$$(f circ g)(x)$$

My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:

Since $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.

Combining these two restrictions, my solution for the range is

$$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$

The given solution, however, is:

$$y in mathbbR mid y neq > 0 $$

I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?

algebra-precalculus functions

share|cite|improve this question

edited Apr 28 at 23:38

Calculemus

asked Apr 28 at 23:20

CalculemusCalculemus

439317

$endgroup$

add a comment | 

$begingroup$

I am attempting a Functions and Inverses self-test offered by the University of Toronto, and I'm trying to understand why my answer for question (4) differs from the given one.

Given $f(x) = dfrac1x - 3$ and $g(x) = sqrtx$, we are asked to find the domain and range of the combined function
$$(f circ g)(x)$$

My solution for the domain matches the given one, and I won't bother reproducing it here, but my solution for the range does not. This is how I determined the range:

Since $(f circ g)(x) = f(g(x)) = dfrac1sqrtx - 3$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrtx geq 0$, which in turn implies that $y geq - dfrac13$.

Combining these two restrictions, my solution for the range is

$$y in mathbbR mid y geq - dfrac 13 wedge y neq 0 $$

The given solution, however, is:

$$y in mathbbR mid y neq > 0 $$

I'm not sure what the $neq >$ notation means. I'm assuming it's a typo, and it's actually supposed to be just a $neq$ sign. But even so, why isn't the $y geq -dfrac13$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?

algebra-precalculus functions

share|cite|improve this question

edited Apr 28 at 23:38

Calculemus

asked Apr 28 at 23:20

CalculemusCalculemus

439317

$endgroup$

share|cite|improve this question

edited Apr 28 at 23:38

Calculemus

asked Apr 28 at 23:20

CalculemusCalculemus

439317

share|cite|improve this question

edited Apr 28 at 23:38

Calculemus

asked Apr 28 at 23:20

CalculemusCalculemus

439317

asked Apr 28 at 23:20

CalculemusCalculemus

439317

asked Apr 28 at 23:20

CalculemusCalculemus

439317

1 Answer
1

active

oldest

votes

4

$begingroup$

The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as $frac 1 t-3: t geq 0, t neq 3$. Find $frac 1 t-3:0 leq t < 3$ and $frac 1 t-3: 3 < t <infty)$ separately. These can be written as $frac 1 s:-3 leq s < 0$ and $frac 1 s: 0 < s <infty)$. Can you compute the range now?

share|cite|improve this answer

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an interesting way to think about it, but I can't follow it entirely. I don't understand where $- infty$ comes from. I don't know how to obtain those intervals from the last two sets. (The method for determining the range of a composite function on the U of T website seems quite different from what's being used here.)
  $endgroup$
  – Calculemus
  Apr 29 at 3:26
  

  
  
  
  

add a comment | 

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4

$begingroup$

The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as $frac 1 t-3: t geq 0, t neq 3$. Find $frac 1 t-3:0 leq t < 3$ and $frac 1 t-3: 3 < t <infty)$ separately. These can be written as $frac 1 s:-3 leq s < 0$ and $frac 1 s: 0 < s <infty)$. Can you compute the range now?

share|cite|improve this answer

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an interesting way to think about it, but I can't follow it entirely. I don't understand where $- infty$ comes from. I don't know how to obtain those intervals from the last two sets. (The method for determining the range of a composite function on the U of T website seems quite different from what's being used here.)
  $endgroup$
  – Calculemus
  Apr 29 at 3:26
  

  
  
  
  

add a comment | 

4

$begingroup$

The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as $frac 1 t-3: t geq 0, t neq 3$. Find $frac 1 t-3:0 leq t < 3$ and $frac 1 t-3: 3 < t <infty)$ separately. These can be written as $frac 1 s:-3 leq s < 0$ and $frac 1 s: 0 < s <infty)$. Can you compute the range now?

share|cite|improve this answer

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is an interesting way to think about it, but I can't follow it entirely. I don't understand where $- infty$ comes from. I don't know how to obtain those intervals from the last two sets. (The method for determining the range of a composite function on the U of T website seems quite different from what's being used here.)
  $endgroup$
  – Calculemus
  Apr 29 at 3:26
  

  
  
  
  

add a comment | 

$begingroup$

The range is $(0,infty) cup (-infty, -frac 1 3]$. To see this write the range as $frac 1 t-3: t geq 0, t neq 3$. Find $frac 1 t-3:0 leq t < 3$ and $frac 1 t-3: 3 < t <infty)$ separately. These can be written as $frac 1 s:-3 leq s < 0$ and $frac 1 s: 0 < s <infty)$. Can you compute the range now?

share|cite|improve this answer

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673

$endgroup$

share|cite|improve this answer

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673

answered Apr 28 at 23:40

Kavi Rama MurthyKavi Rama Murthy

79.5k53673