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4

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 $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, 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 $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.

Combining these two restrictions, my solution for the range is

$${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$

The given solution, however, is:

$${y in mathbb{R} mid y neq > 0 }$$

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 $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?

algebra-precalculus functions

share|cite|improve this question

edited 2 hours ago

Calculemus

asked 2 hours ago

CalculemusCalculemus

432317

$endgroup$

add a comment | 

4

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 $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, 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 $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.

Combining these two restrictions, my solution for the range is

$${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$

The given solution, however, is:

$${y in mathbb{R} mid y neq > 0 }$$

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 $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?

algebra-precalculus functions

share|cite|improve this question

edited 2 hours ago

Calculemus

asked 2 hours ago

CalculemusCalculemus

432317

$endgroup$

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 $f(x) = dfrac{1}{x - 3}$ and $g(x) = sqrt{x}$, 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 $(f circ g)(x) = f(g(x)) = dfrac{1}{sqrt{x} - 3}$, it's easy to see that $y neq 0$, since the numerator isn't $0$. We also know that $sqrt{x} geq 0$, which in turn implies that $y geq - dfrac{1}{3}$.

Combining these two restrictions, my solution for the range is

$${y in mathbb{R} mid y geq - dfrac {1}{3} wedge y neq 0 }$$

The given solution, however, is:

$${y in mathbb{R} mid y neq > 0 }$$

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 $y geq -dfrac{1}{3}$ restriction mentioned? Was I wrong in concluding it? Is it optional to mention it?

algebra-precalculus functions

share|cite|improve this question

edited 2 hours ago

Calculemus

asked 2 hours ago

CalculemusCalculemus

432317

$endgroup$

share|cite|improve this question

edited 2 hours ago

Calculemus

asked 2 hours ago

CalculemusCalculemus

432317

share|cite|improve this question

edited 2 hours ago

Calculemus

asked 2 hours ago

CalculemusCalculemus

432317

asked 2 hours ago

CalculemusCalculemus

432317

asked 2 hours ago

CalculemusCalculemus

432317

1 Answer
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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 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572

$endgroup$

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 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572

$endgroup$

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 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572

$endgroup$

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 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572

answered 2 hours ago

Kavi Rama MurthyKavi Rama Murthy

78.5k53572