Mixing
How to find the range of a function without a graph?
$begingroup$
Okay so I know how to find the domain of the function $f(x) = frac{3x^2}{x^2-1}$, which is $xneq 1$ and $xneq -1$, but I'm totally confused on how to find the range without using a graph.
calculus functions
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
$endgroup$
add a comment |
$begingroup$
Okay so I know how to find the domain of the function $f(x) = frac{3x^2}{x^2-1}$, which is $xneq 1$ and $xneq -1$, but I'm totally confused on how to find the range without using a graph.
calculus functions
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
$endgroup$
1
$begingroup$
Hint: most meaningful part is the behavior near the singularities. what happens if $x$ is slightly greater than $1$? What if it is slightly less than $1$?
$endgroup$
– lulu
Sep 25 '15 at 20:26
$begingroup$
Check for horizonal and slant asymptotes, and continuity.
$endgroup$
– Transcendental
Sep 25 '15 at 20:27
$begingroup$
Making a sign chart for the function and for its derivative will be quite helpful. Those will essentially let you sketch the graph, so they furnish the same information.
$endgroup$
– MPW
Sep 25 '15 at 20:29
1
$begingroup$
Note that it is even, so you only need to consider $xge 0$ (and $x neq 1$). Note that $f(0) = 0$. If $x in (0,1)$ the denominator is negative, and $lim_{x uparrow 1} f(x) = -infty$. Hence by continuity we see that the range contains $(-infty,0]$. Note that we can write $f(x) = { 3 over 1-{1 over x^2}}$, so we can see that for $x>1$, the value of $f(x) $ lies in $(3, infty)$.
$endgroup$
– copper.hat
Sep 25 '15 at 20:32
add a comment |
$begingroup$
Okay so I know how to find the domain of the function $f(x) = frac{3x^2}{x^2-1}$, which is $xneq 1$ and $xneq -1$, but I'm totally confused on how to find the range without using a graph.
calculus functions
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
$endgroup$
Okay so I know how to find the domain of the function $f(x) = frac{3x^2}{x^2-1}$, which is $xneq 1$ and $xneq -1$, but I'm totally confused on how to find the range without using a graph.
calculus functions
calculus functions
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
edited Sep 15 '16 at 6:07
Martin Sleziak
44.8k9118272
44.8k9118272
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
asked Sep 25 '15 at 20:23
ppgkingppgking
6112
6112
1
$begingroup$
Hint: most meaningful part is the behavior near the singularities. what happens if $x$ is slightly greater than $1$? What if it is slightly less than $1$?
$endgroup$
– lulu
Sep 25 '15 at 20:26
$begingroup$
Check for horizonal and slant asymptotes, and continuity.
$endgroup$
– Transcendental
Sep 25 '15 at 20:27
$begingroup$
Making a sign chart for the function and for its derivative will be quite helpful. Those will essentially let you sketch the graph, so they furnish the same information.
$endgroup$
– MPW
Sep 25 '15 at 20:29
1
$begingroup$
Note that it is even, so you only need to consider $xge 0$ (and $x neq 1$). Note that $f(0) = 0$. If $x in (0,1)$ the denominator is negative, and $lim_{x uparrow 1} f(x) = -infty$. Hence by continuity we see that the range contains $(-infty,0]$. Note that we can write $f(x) = { 3 over 1-{1 over x^2}}$, so we can see that for $x>1$, the value of $f(x) $ lies in $(3, infty)$.
$endgroup$
– copper.hat
Sep 25 '15 at 20:32
add a comment |
1
$begingroup$
Hint: most meaningful part is the behavior near the singularities. what happens if $x$ is slightly greater than $1$? What if it is slightly less than $1$?
$endgroup$
– lulu
Sep 25 '15 at 20:26
$begingroup$
Check for horizonal and slant asymptotes, and continuity.
$endgroup$
– Transcendental
Sep 25 '15 at 20:27
$begingroup$
Making a sign chart for the function and for its derivative will be quite helpful. Those will essentially let you sketch the graph, so they furnish the same information.
$endgroup$
– MPW
Sep 25 '15 at 20:29
1
$begingroup$
Note that it is even, so you only need to consider $xge 0$ (and $x neq 1$). Note that $f(0) = 0$. If $x in (0,1)$ the denominator is negative, and $lim_{x uparrow 1} f(x) = -infty$. Hence by continuity we see that the range contains $(-infty,0]$. Note that we can write $f(x) = { 3 over 1-{1 over x^2}}$, so we can see that for $x>1$, the value of $f(x) $ lies in $(3, infty)$.
$endgroup$
– copper.hat
Sep 25 '15 at 20:32
1
1
$begingroup$
Hint: most meaningful part is the behavior near the singularities. what happens if $x$ is slightly greater than $1$? What if it is slightly less than $1$?
$endgroup$
– lulu
Sep 25 '15 at 20:26
$begingroup$
Hint: most meaningful part is the behavior near the singularities. what happens if $x$ is slightly greater than $1$? What if it is slightly less than $1$?
$endgroup$
– lulu
Sep 25 '15 at 20:26
$begingroup$
Check for horizonal and slant asymptotes, and continuity.
$endgroup$
– Transcendental
Sep 25 '15 at 20:27
$begingroup$
Check for horizonal and slant asymptotes, and continuity.
$endgroup$
– Transcendental
Sep 25 '15 at 20:27
$begingroup$
Making a sign chart for the function and for its derivative will be quite helpful. Those will essentially let you sketch the graph, so they furnish the same information.
$endgroup$
– MPW
Sep 25 '15 at 20:29
$begingroup$
Making a sign chart for the function and for its derivative will be quite helpful. Those will essentially let you sketch the graph, so they furnish the same information.
$endgroup$
– MPW
Sep 25 '15 at 20:29
1
1
$begingroup$
Note that it is even, so you only need to consider $xge 0$ (and $x neq 1$). Note that $f(0) = 0$. If $x in (0,1)$ the denominator is negative, and $lim_{x uparrow 1} f(x) = -infty$. Hence by continuity we see that the range contains $(-infty,0]$. Note that we can write $f(x) = { 3 over 1-{1 over x^2}}$, so we can see that for $x>1$, the value of $f(x) $ lies in $(3, infty)$.
$endgroup$
– copper.hat
Sep 25 '15 at 20:32
$begingroup$
Note that it is even, so you only need to consider $xge 0$ (and $x neq 1$). Note that $f(0) = 0$. If $x in (0,1)$ the denominator is negative, and $lim_{x uparrow 1} f(x) = -infty$. Hence by continuity we see that the range contains $(-infty,0]$. Note that we can write $f(x) = { 3 over 1-{1 over x^2}}$, so we can see that for $x>1$, the value of $f(x) $ lies in $(3, infty)$.
$endgroup$
– copper.hat
Sep 25 '15 at 20:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Set $$y=frac{3x^2}{x^2-1}$$ and rearrange to get $$(y-3)x^2-y=0$$
This quadratic has real roots provided $b^2-4acgeq 0$ which translates as $$y(y-3)geq 0$$
Noting that $y=3$ is the horizontal asymptote, the solution set, i.e.the range, is $$yleq 0, y>3$$
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
$endgroup$
add a comment |
$begingroup$
Hint:
find for what values of $k$ the equation $k=dfrac{3x^2}{x^2-1}$ has real solutions.
For $ xne pm 1$ you have $ x^2= dfrac{k}{k-3}$ so this fraction must be not negative: $dfrac{k}{k-3}ge 0$. The solution is the range.
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
$endgroup$
add a comment |
$begingroup$
Write the function as
$$f(x)=frac{3x^2-3+3}{x^2-1}=3+frac3{x^2-1}$$
As $f$ is even, we may suppose $xge 0,enspace xne q 1$.
Now the range of $x^2-1$ is $[-1,+infty)$ and for $f(x)$, the value $x^2=1$ is excluded, hence the range of $;dfrac3{x^2-1}$ is $(-infty,-3] $ for $xin [0,1)$ and $(0,+infty)$ for $x>1$.
Hence the range of $f(x)=3+dfrac3{x^2-1}$ is $;(-infty,0]cup (3,+infty)$.
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Set $$y=frac{3x^2}{x^2-1}$$ and rearrange to get $$(y-3)x^2-y=0$$
This quadratic has real roots provided $b^2-4acgeq 0$ which translates as $$y(y-3)geq 0$$
Noting that $y=3$ is the horizontal asymptote, the solution set, i.e.the range, is $$yleq 0, y>3$$
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
$endgroup$
add a comment |
$begingroup$
Set $$y=frac{3x^2}{x^2-1}$$ and rearrange to get $$(y-3)x^2-y=0$$
This quadratic has real roots provided $b^2-4acgeq 0$ which translates as $$y(y-3)geq 0$$
Noting that $y=3$ is the horizontal asymptote, the solution set, i.e.the range, is $$yleq 0, y>3$$
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
$endgroup$
add a comment |
$begingroup$
Set $$y=frac{3x^2}{x^2-1}$$ and rearrange to get $$(y-3)x^2-y=0$$
This quadratic has real roots provided $b^2-4acgeq 0$ which translates as $$y(y-3)geq 0$$
Noting that $y=3$ is the horizontal asymptote, the solution set, i.e.the range, is $$yleq 0, y>3$$
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
$endgroup$
Set $$y=frac{3x^2}{x^2-1}$$ and rearrange to get $$(y-3)x^2-y=0$$
This quadratic has real roots provided $b^2-4acgeq 0$ which translates as $$y(y-3)geq 0$$
Noting that $y=3$ is the horizontal asymptote, the solution set, i.e.the range, is $$yleq 0, y>3$$
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
answered Sep 25 '15 at 20:42
David QuinnDavid Quinn
24k21141
24k21141
add a comment |
add a comment |
$begingroup$
Hint:
find for what values of $k$ the equation $k=dfrac{3x^2}{x^2-1}$ has real solutions.
For $ xne pm 1$ you have $ x^2= dfrac{k}{k-3}$ so this fraction must be not negative: $dfrac{k}{k-3}ge 0$. The solution is the range.
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
$endgroup$
add a comment |
$begingroup$
Hint:
find for what values of $k$ the equation $k=dfrac{3x^2}{x^2-1}$ has real solutions.
For $ xne pm 1$ you have $ x^2= dfrac{k}{k-3}$ so this fraction must be not negative: $dfrac{k}{k-3}ge 0$. The solution is the range.
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
$endgroup$
add a comment |
$begingroup$
Hint:
find for what values of $k$ the equation $k=dfrac{3x^2}{x^2-1}$ has real solutions.
For $ xne pm 1$ you have $ x^2= dfrac{k}{k-3}$ so this fraction must be not negative: $dfrac{k}{k-3}ge 0$. The solution is the range.
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
$endgroup$
Hint:
find for what values of $k$ the equation $k=dfrac{3x^2}{x^2-1}$ has real solutions.
For $ xne pm 1$ you have $ x^2= dfrac{k}{k-3}$ so this fraction must be not negative: $dfrac{k}{k-3}ge 0$. The solution is the range.
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
answered Sep 25 '15 at 20:41
Emilio NovatiEmilio Novati
52k43474
52k43474
add a comment |
add a comment |
$begingroup$
Write the function as
$$f(x)=frac{3x^2-3+3}{x^2-1}=3+frac3{x^2-1}$$
As $f$ is even, we may suppose $xge 0,enspace xne q 1$.
Now the range of $x^2-1$ is $[-1,+infty)$ and for $f(x)$, the value $x^2=1$ is excluded, hence the range of $;dfrac3{x^2-1}$ is $(-infty,-3] $ for $xin [0,1)$ and $(0,+infty)$ for $x>1$.
Hence the range of $f(x)=3+dfrac3{x^2-1}$ is $;(-infty,0]cup (3,+infty)$.
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
$endgroup$
add a comment |
$begingroup$
Write the function as
$$f(x)=frac{3x^2-3+3}{x^2-1}=3+frac3{x^2-1}$$
As $f$ is even, we may suppose $xge 0,enspace xne q 1$.
Now the range of $x^2-1$ is $[-1,+infty)$ and for $f(x)$, the value $x^2=1$ is excluded, hence the range of $;dfrac3{x^2-1}$ is $(-infty,-3] $ for $xin [0,1)$ and $(0,+infty)$ for $x>1$.
Hence the range of $f(x)=3+dfrac3{x^2-1}$ is $;(-infty,0]cup (3,+infty)$.
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
$endgroup$
add a comment |
$begingroup$
Write the function as
$$f(x)=frac{3x^2-3+3}{x^2-1}=3+frac3{x^2-1}$$
As $f$ is even, we may suppose $xge 0,enspace xne q 1$.
Now the range of $x^2-1$ is $[-1,+infty)$ and for $f(x)$, the value $x^2=1$ is excluded, hence the range of $;dfrac3{x^2-1}$ is $(-infty,-3] $ for $xin [0,1)$ and $(0,+infty)$ for $x>1$.
Hence the range of $f(x)=3+dfrac3{x^2-1}$ is $;(-infty,0]cup (3,+infty)$.
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
$endgroup$
Write the function as
$$f(x)=frac{3x^2-3+3}{x^2-1}=3+frac3{x^2-1}$$
As $f$ is even, we may suppose $xge 0,enspace xne q 1$.
Now the range of $x^2-1$ is $[-1,+infty)$ and for $f(x)$, the value $x^2=1$ is excluded, hence the range of $;dfrac3{x^2-1}$ is $(-infty,-3] $ for $xin [0,1)$ and $(0,+infty)$ for $x>1$.
Hence the range of $f(x)=3+dfrac3{x^2-1}$ is $;(-infty,0]cup (3,+infty)$.
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
edited Jan 9 at 22:35
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
answered Sep 25 '15 at 20:40
BernardBernard
120k740113
120k740113
add a comment |
add a comment |
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1
$begingroup$
Hint: most meaningful part is the behavior near the singularities. what happens if $x$ is slightly greater than $1$? What if it is slightly less than $1$?
$endgroup$
– lulu
Sep 25 '15 at 20:26
$begingroup$
Check for horizonal and slant asymptotes, and continuity.
$endgroup$
– Transcendental
Sep 25 '15 at 20:27
$begingroup$
Making a sign chart for the function and for its derivative will be quite helpful. Those will essentially let you sketch the graph, so they furnish the same information.
$endgroup$
– MPW
Sep 25 '15 at 20:29
1
$begingroup$
Note that it is even, so you only need to consider $xge 0$ (and $x neq 1$). Note that $f(0) = 0$. If $x in (0,1)$ the denominator is negative, and $lim_{x uparrow 1} f(x) = -infty$. Hence by continuity we see that the range contains $(-infty,0]$. Note that we can write $f(x) = { 3 over 1-{1 over x^2}}$, so we can see that for $x>1$, the value of $f(x) $ lies in $(3, infty)$.
$endgroup$
– copper.hat
Sep 25 '15 at 20:32