Finding the range of a function without graph?
$begingroup$
I'm asked to find the range of the function :
$
(-x^2 , +1)
$
Taking R as a real number which represents a quantity along a line and graphing :
then the range for this function from viewing the graph appears to be :
$
R = (-infty , +infty)
$
Is there an alternative method of finding the range of this function without using a graph ?
Watching the khan academy tutorial suggests using graphs : https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/domain-and-range-from-graphs but is there a pure algebraic method instead of using graphs ?
It is not clear what the range is when the range appears infinite as how do we know that at some point on the axis the range functions stops tending towards infinity ?
algebra-precalculus functions
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
$endgroup$
add a comment |
$begingroup$
I'm asked to find the range of the function :
$
(-x^2 , +1)
$
Taking R as a real number which represents a quantity along a line and graphing :
then the range for this function from viewing the graph appears to be :
$
R = (-infty , +infty)
$
Is there an alternative method of finding the range of this function without using a graph ?
Watching the khan academy tutorial suggests using graphs : https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/domain-and-range-from-graphs but is there a pure algebraic method instead of using graphs ?
It is not clear what the range is when the range appears infinite as how do we know that at some point on the axis the range functions stops tending towards infinity ?
algebra-precalculus functions
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
$endgroup$
add a comment |
$begingroup$
I'm asked to find the range of the function :
$
(-x^2 , +1)
$
Taking R as a real number which represents a quantity along a line and graphing :
then the range for this function from viewing the graph appears to be :
$
R = (-infty , +infty)
$
Is there an alternative method of finding the range of this function without using a graph ?
Watching the khan academy tutorial suggests using graphs : https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/domain-and-range-from-graphs but is there a pure algebraic method instead of using graphs ?
It is not clear what the range is when the range appears infinite as how do we know that at some point on the axis the range functions stops tending towards infinity ?
algebra-precalculus functions
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
$endgroup$
I'm asked to find the range of the function :
$
(-x^2 , +1)
$
Taking R as a real number which represents a quantity along a line and graphing :
then the range for this function from viewing the graph appears to be :
$
R = (-infty , +infty)
$
Is there an alternative method of finding the range of this function without using a graph ?
Watching the khan academy tutorial suggests using graphs : https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/domain-and-range-from-graphs but is there a pure algebraic method instead of using graphs ?
It is not clear what the range is when the range appears infinite as how do we know that at some point on the axis the range functions stops tending towards infinity ?
algebra-precalculus functions
algebra-precalculus functions
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
edited Sep 24 '16 at 11:10
Alex M.
28.6k103259
28.6k103259
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
asked Sep 24 '16 at 8:53
blue-skyblue-sky
1,03311322
1,03311322
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Notice that $x^2 ge 0$ for all $x in Bbb R$, which means that $1 - x^2 le 1$ for all $x in Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $infty$, therefore $1-x^2$ will decrease towards $-infty$. Finally, notice that $x^2$ is surjective (for every $y ge 0$, there exist $sqrt y$). Therefore, the values of $f$ lay between $- infty$ and $1$, meaning that the range is $(-infty, 1]$.
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
$endgroup$
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
1
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
add a comment |
$begingroup$
You see that if we put x as infinity we get f(x)=-∞.
Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing.
$frac{d}{dx}(-x^2 + 1) = -2x$
$-2x=0 =>$ f(x) is maximum when x=0
Therefore maximum value of f(x) is +1
$f(x)epsilon (-∞,+1) $ here is the max value on graph
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that $x^2 ge 0$ for all $x in Bbb R$, which means that $1 - x^2 le 1$ for all $x in Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $infty$, therefore $1-x^2$ will decrease towards $-infty$. Finally, notice that $x^2$ is surjective (for every $y ge 0$, there exist $sqrt y$). Therefore, the values of $f$ lay between $- infty$ and $1$, meaning that the range is $(-infty, 1]$.
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
$endgroup$
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
1
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
add a comment |
$begingroup$
Notice that $x^2 ge 0$ for all $x in Bbb R$, which means that $1 - x^2 le 1$ for all $x in Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $infty$, therefore $1-x^2$ will decrease towards $-infty$. Finally, notice that $x^2$ is surjective (for every $y ge 0$, there exist $sqrt y$). Therefore, the values of $f$ lay between $- infty$ and $1$, meaning that the range is $(-infty, 1]$.
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
$endgroup$
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
1
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
add a comment |
$begingroup$
Notice that $x^2 ge 0$ for all $x in Bbb R$, which means that $1 - x^2 le 1$ for all $x in Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $infty$, therefore $1-x^2$ will decrease towards $-infty$. Finally, notice that $x^2$ is surjective (for every $y ge 0$, there exist $sqrt y$). Therefore, the values of $f$ lay between $- infty$ and $1$, meaning that the range is $(-infty, 1]$.
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
$endgroup$
Notice that $x^2 ge 0$ for all $x in Bbb R$, which means that $1 - x^2 le 1$ for all $x in Bbb R$, with equality for $x=0$. Notice, too, that $x^2$ increases towards $infty$, therefore $1-x^2$ will decrease towards $-infty$. Finally, notice that $x^2$ is surjective (for every $y ge 0$, there exist $sqrt y$). Therefore, the values of $f$ lay between $- infty$ and $1$, meaning that the range is $(-infty, 1]$.
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
answered Sep 24 '16 at 10:13
Alex M.Alex M.
28.6k103259
28.6k103259
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
1
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
add a comment |
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
1
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
Small addendum, because that is a bit unclear in your answer: $x^2$ is surjective as a function $mathbb{R}to[0,infty)$, but not as a function $mathbb{R}tomathbb{R}$.
$endgroup$
– Janik
Sep 24 '16 at 10:25
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@Janik: I've written "for every $sqrt y ge 0$", so this should say it all.
$endgroup$
– Alex M.
Sep 24 '16 at 10:35
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
$begingroup$
@AlexM. $1-x^2=0$ , can you elaborate how you arrive at $1-x^2<=1$ from this assertion ?
$endgroup$
– blue-sky
Sep 24 '16 at 10:50
1
1
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
$begingroup$
@blue-sky: I don't get your question, you must have misunderstood me. I'm saying that $x^2 ge 0 iff x^2 - 1 ge -1 iff -x^2 + 1 le 1$ (for the last inequality multiply the preceding one by $-1$).
$endgroup$
– Alex M.
Sep 24 '16 at 11:00
add a comment |
$begingroup$
You see that if we put x as infinity we get f(x)=-∞.
Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing.
$frac{d}{dx}(-x^2 + 1) = -2x$
$-2x=0 =>$ f(x) is maximum when x=0
Therefore maximum value of f(x) is +1
$f(x)epsilon (-∞,+1) $ here is the max value on graph
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
$endgroup$
add a comment |
$begingroup$
You see that if we put x as infinity we get f(x)=-∞.
Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing.
$frac{d}{dx}(-x^2 + 1) = -2x$
$-2x=0 =>$ f(x) is maximum when x=0
Therefore maximum value of f(x) is +1
$f(x)epsilon (-∞,+1) $ here is the max value on graph
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
$endgroup$
add a comment |
$begingroup$
You see that if we put x as infinity we get f(x)=-∞.
Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing.
$frac{d}{dx}(-x^2 + 1) = -2x$
$-2x=0 =>$ f(x) is maximum when x=0
Therefore maximum value of f(x) is +1
$f(x)epsilon (-∞,+1) $ here is the max value on graph
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
$endgroup$
You see that if we put x as infinity we get f(x)=-∞.
Now to find the maximum value of functions just differentiate it and put the derivative equal to zero to find the point where the function stops increasing and starts decreasing.
$frac{d}{dx}(-x^2 + 1) = -2x$
$-2x=0 =>$ f(x) is maximum when x=0
Therefore maximum value of f(x) is +1
$f(x)epsilon (-∞,+1) $ here is the max value on graph
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
edited Jan 7 at 14:19
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
answered Jan 6 at 12:47
Swap NayakSwap Nayak
72
72
add a comment |
add a comment |
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