Range of a rational function with radicals
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Find the range of the function
$$frac65sqrtx^2-10x+29 - 2$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$frac65sqrtx^2-10x+29 - 2 = y$
getting the inverse,
$frac65sqrty^2-10y+29 - 2 = x$
$frac4x^2+24x+3625x^2= y^2-10y +29$
Then it would be a quadratic function in y, but the discriminant becomes really big
$100- 116(frac4x^2+24x+3625x^2)$
functions
asked Aug 19 at 9:41
SuperMage1
722210
add a comment |Â
up vote
2
down vote
favorite
Find the range of the function
$$frac65sqrtx^2-10x+29 - 2$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$frac65sqrtx^2-10x+29 - 2 = y$
getting the inverse,
$frac65sqrty^2-10y+29 - 2 = x$
$frac4x^2+24x+3625x^2= y^2-10y +29$
Then it would be a quadratic function in y, but the discriminant becomes really big
$100- 116(frac4x^2+24x+3625x^2)$
functions
asked Aug 19 at 9:41
SuperMage1
722210
What do you mean by "I tried using inverses"? Could you show your work?
– A. Pongrácz
Aug 19 at 9:43
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum.
– Bruce
Aug 19 at 9:49
Completing the square even saves you the calculus
– Bruce
Aug 19 at 9:52
You can also write $ y^2-10y +29=left ( 1 over 5left ( 6over x+2right ) right )^2$
– N74
Aug 19 at 17:05
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the range of the function
$$frac65sqrtx^2-10x+29 - 2$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$frac65sqrtx^2-10x+29 - 2 = y$
getting the inverse,
$frac65sqrty^2-10y+29 - 2 = x$
$frac4x^2+24x+3625x^2= y^2-10y +29$
Then it would be a quadratic function in y, but the discriminant becomes really big
$100- 116(frac4x^2+24x+3625x^2)$
functions
asked Aug 19 at 9:41
SuperMage1
722210
Find the range of the function
$$frac65sqrtx^2-10x+29 - 2$$
I tried using inverses, but the equation got super messy and I dont think its a good method for this problem.
$frac65sqrtx^2-10x+29 - 2 = y$
getting the inverse,
$frac65sqrty^2-10y+29 - 2 = x$
$frac4x^2+24x+3625x^2= y^2-10y +29$
Then it would be a quadratic function in y, but the discriminant becomes really big
$100- 116(frac4x^2+24x+3625x^2)$
functions
functions
asked Aug 19 at 9:41
SuperMage1
722210
asked Aug 19 at 9:41
SuperMage1
722210
edited Aug 19 at 9:50
asked Aug 19 at 9:41
SuperMage1
722210
asked Aug 19 at 9:41
SuperMage1
722210
asked Aug 19 at 9:41
SuperMage1
722210
722210
What do you mean by "I tried using inverses"? Could you show your work?
– A. Pongrácz
Aug 19 at 9:43
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum.
– Bruce
Aug 19 at 9:49
Completing the square even saves you the calculus
– Bruce
Aug 19 at 9:52
You can also write $ y^2-10y +29=left ( 1 over 5left ( 6over x+2right ) right )^2$
– N74
Aug 19 at 17:05
add a comment |Â
What do you mean by "I tried using inverses"? Could you show your work?
– A. Pongrácz
Aug 19 at 9:43
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum.
– Bruce
Aug 19 at 9:49
Completing the square even saves you the calculus
– Bruce
Aug 19 at 9:52
You can also write $ y^2-10y +29=left ( 1 over 5left ( 6over x+2right ) right )^2$
– N74
Aug 19 at 17:05
What do you mean by "I tried using inverses"? Could you show your work?
– A. Pongrácz
Aug 19 at 9:43
What do you mean by "I tried using inverses"? Could you show your work?
– A. Pongrácz
Aug 19 at 9:43
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum.
– Bruce
Aug 19 at 9:49
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum.
– Bruce
Aug 19 at 9:49
Completing the square even saves you the calculus
– Bruce
Aug 19 at 9:52
Completing the square even saves you the calculus
– Bruce
Aug 19 at 9:52
You can also write $ y^2-10y +29=left ( 1 over 5left ( 6over x+2right ) right )^2$
– N74
Aug 19 at 17:05
You can also write $ y^2-10y +29=left ( 1 over 5left ( 6over x+2right ) right )^2$
– N74
Aug 19 at 17:05
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
$frac65sqrt(x^2-10x+29)-2=frac65sqrt((x-5)^2+4)-2$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $frac34$ and you have a range $]0,frac34]$
answered Aug 19 at 9:54
Bruce
520113
add a comment |Â
up vote
7
down vote
Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(x-5)^2+4 in [4,+infty)$$
$$sqrt(x-5)^2+4 in [2,+infty)$$
$$5sqrt(x-5)^2+4-2 in [8,+infty)$$
Does that help?
answered Aug 19 at 9:53
StackTD
20.4k1544
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$frac65sqrt(x^2-10x+29)-2=frac65sqrt((x-5)^2+4)-2$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $frac34$ and you have a range $]0,frac34]$
answered Aug 19 at 9:54
Bruce
520113
add a comment |Â
up vote
2
down vote
accepted
$frac65sqrt(x^2-10x+29)-2=frac65sqrt((x-5)^2+4)-2$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $frac34$ and you have a range $]0,frac34]$
answered Aug 19 at 9:54
Bruce
520113
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$frac65sqrt(x^2-10x+29)-2=frac65sqrt((x-5)^2+4)-2$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $frac34$ and you have a range $]0,frac34]$
answered Aug 19 at 9:54
Bruce
520113
$frac65sqrt(x^2-10x+29)-2=frac65sqrt((x-5)^2+4)-2$
so you can see that there will be a maximum at $x=5$ and as $x$ tends to either positive or negative infinity the graph goes to zero.
So just plug in $x=5$ to get $frac34$ and you have a range $]0,frac34]$
answered Aug 19 at 9:54
Bruce
520113
answered Aug 19 at 9:54
Bruce
520113
answered Aug 19 at 9:54
Bruce
520113
answered Aug 19 at 9:54
Bruce
520113
520113
add a comment |Â
add a comment |Â
up vote
7
down vote
Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(x-5)^2+4 in [4,+infty)$$
$$sqrt(x-5)^2+4 in [2,+infty)$$
$$5sqrt(x-5)^2+4-2 in [8,+infty)$$
Does that help?
answered Aug 19 at 9:53
StackTD
20.4k1544
add a comment |Â
up vote
7
down vote
Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(x-5)^2+4 in [4,+infty)$$
$$sqrt(x-5)^2+4 in [2,+infty)$$
$$5sqrt(x-5)^2+4-2 in [8,+infty)$$
Does that help?
answered Aug 19 at 9:53
StackTD
20.4k1544
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(x-5)^2+4 in [4,+infty)$$
$$sqrt(x-5)^2+4 in [2,+infty)$$
$$5sqrt(x-5)^2+4-2 in [8,+infty)$$
Does that help?
answered Aug 19 at 9:53
StackTD
20.4k1544
Hint
If you rewrite $x^2-10x+29=(x-5)^2+4$, it's easier to see that:
$$(x-5)^2+4 in [4,+infty)$$
$$sqrt(x-5)^2+4 in [2,+infty)$$
$$5sqrt(x-5)^2+4-2 in [8,+infty)$$
Does that help?
answered Aug 19 at 9:53
StackTD
20.4k1544
answered Aug 19 at 9:53
StackTD
20.4k1544
answered Aug 19 at 9:53
StackTD
20.4k1544
answered Aug 19 at 9:53
StackTD
20.4k1544
20.4k1544
add a comment |Â
add a comment |Â
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What do you mean by "I tried using inverses"? Could you show your work?
– A. Pongrácz
Aug 19 at 9:43
A graph will show you some things that will help. It isn't hard to see that it can't be negative and a little calculus will help you find the single maximum.
– Bruce
Aug 19 at 9:49
Completing the square even saves you the calculus
– Bruce
Aug 19 at 9:52
You can also write $ y^2-10y +29=left ( 1 over 5left ( 6over x+2right ) right )^2$
– N74
Aug 19 at 17:05