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What is the range of this combined function?
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What is the range of this combined function?
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$begingroup$
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
asked 2 hours ago
CalculemusCalculemus
427317
$endgroup$
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
asked 2 hours ago
CalculemusCalculemus
427317
$endgroup$
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
asked 2 hours ago
CalculemusCalculemus
427317
$endgroup$
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
algebra-precalculus functions
asked 2 hours ago
CalculemusCalculemus
427317
asked 2 hours ago
CalculemusCalculemus
427317
edited 2 hours ago
Calculemus
asked 2 hours ago
CalculemusCalculemus
427317
asked 2 hours ago
CalculemusCalculemus
427317
asked 2 hours ago
CalculemusCalculemus
427317
427317
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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?
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
78.5k53572
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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?
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?
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?
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
78.5k53572
$endgroup$
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?
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
answered 2 hours ago
Kavi Rama MurthyKavi Rama Murthy
78.5k53572
78.5k53572
add a comment |
add a comment |
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