Mixing
How to find the range of $y=frac{x^2+1}{x+1}$ without using derivative?
$begingroup$
The only thing I know with this equation is $y=frac{x^2+1}{x+1}=x+1-frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
inequality a.m.-g.m.-inequality
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
$endgroup$
add a comment |
$begingroup$
The only thing I know with this equation is $y=frac{x^2+1}{x+1}=x+1-frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
inequality a.m.-g.m.-inequality
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
$endgroup$
$begingroup$
math.stackexchange.com/questions/174905/…
$endgroup$
– lab bhattacharjee
Jan 14 at 12:40
add a comment |
$begingroup$
The only thing I know with this equation is $y=frac{x^2+1}{x+1}=x+1-frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
inequality a.m.-g.m.-inequality
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
$endgroup$
The only thing I know with this equation is $y=frac{x^2+1}{x+1}=x+1-frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
inequality a.m.-g.m.-inequality
inequality a.m.-g.m.-inequality
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
edited Jan 14 at 8:57
Michael Rozenberg
103k1891195
103k1891195
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
asked Jan 14 at 8:44
yuanming luoyuanming luo
737
737
$begingroup$
math.stackexchange.com/questions/174905/…
$endgroup$
– lab bhattacharjee
Jan 14 at 12:40
add a comment |
$begingroup$
math.stackexchange.com/questions/174905/…
$endgroup$
– lab bhattacharjee
Jan 14 at 12:40
$begingroup$
math.stackexchange.com/questions/174905/…
$endgroup$
– lab bhattacharjee
Jan 14 at 12:40
$begingroup$
math.stackexchange.com/questions/174905/…
$endgroup$
– lab bhattacharjee
Jan 14 at 12:40
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Does not exist.
Try $xrightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$frac{x^2+1}{x+1}=frac{x^2-1+2}{x+1}=x-1+frac{2}{x+1}=$$
$$=x+1+frac{2}{x+1}-2geq2sqrt{(x+1)cdotfrac{2}{x+1}}-2=2sqrt2-2.$$
The equality occurs for $x+1=frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2sqrt2-2,+infty).$
For $x<-1$ we can get the range by the similar way.
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
|
show 4 more comments
$begingroup$
The equation $$y={x^2+1over x+1}=ato x^2-ax-(a-1)=0$$for $xne -1$ has a real root if and only if $$Deltage0 iff a^2+4a-4ge 0iff age 2sqrt 2-2text{ or }ale -2sqrt 2-2$$in that case, the root is $ne -1$ automatically (if not, it leads to the impossible non-equality $a-1=a+1$), Therefore the range simply is $$(-infty,-2sqrt 2-2] bigcup [2sqrt 2-2,infty)$$
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=frac{x^2+1}{x+1}$
$$Rightarrow yx + y = x^2 +1 Rightarrow -x^2 + yx +(y-1)=0 Rightarrow x = frac{-y pmsqrt{y^2 + 4(y-1)}}{-2} Rightarrow frac{-y pm sqrt{(y-(-2-sqrt5))(y - (sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = frac{-sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
$endgroup$
add a comment |
$begingroup$
Since $$lim_{xto-infty} frac{x^2+1}{x+1} = -infty,$$
the minimum does not exist.
answered Jan 14 at 8:46
5xum5xum
90.8k394161
$endgroup$
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Does not exist.
Try $xrightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$frac{x^2+1}{x+1}=frac{x^2-1+2}{x+1}=x-1+frac{2}{x+1}=$$
$$=x+1+frac{2}{x+1}-2geq2sqrt{(x+1)cdotfrac{2}{x+1}}-2=2sqrt2-2.$$
The equality occurs for $x+1=frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2sqrt2-2,+infty).$
For $x<-1$ we can get the range by the similar way.
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
|
show 4 more comments
$begingroup$
Does not exist.
Try $xrightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$frac{x^2+1}{x+1}=frac{x^2-1+2}{x+1}=x-1+frac{2}{x+1}=$$
$$=x+1+frac{2}{x+1}-2geq2sqrt{(x+1)cdotfrac{2}{x+1}}-2=2sqrt2-2.$$
The equality occurs for $x+1=frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2sqrt2-2,+infty).$
For $x<-1$ we can get the range by the similar way.
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
|
show 4 more comments
$begingroup$
Does not exist.
Try $xrightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$frac{x^2+1}{x+1}=frac{x^2-1+2}{x+1}=x-1+frac{2}{x+1}=$$
$$=x+1+frac{2}{x+1}-2geq2sqrt{(x+1)cdotfrac{2}{x+1}}-2=2sqrt2-2.$$
The equality occurs for $x+1=frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2sqrt2-2,+infty).$
For $x<-1$ we can get the range by the similar way.
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
$endgroup$
Does not exist.
Try $xrightarrow-1^-$.
For $x>-1$ by AM-GM we obtain:
$$frac{x^2+1}{x+1}=frac{x^2-1+2}{x+1}=x-1+frac{2}{x+1}=$$
$$=x+1+frac{2}{x+1}-2geq2sqrt{(x+1)cdotfrac{2}{x+1}}-2=2sqrt2-2.$$
The equality occurs for $x+1=frac{2}{x+1},$ which says that we got a minimal value and the local minimal value.
Thus, the range for $x>0$ it's $[2sqrt2-2,+infty).$
For $x<-1$ we can get the range by the similar way.
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
edited Jan 14 at 8:55
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
answered Jan 14 at 8:46
Michael RozenbergMichael Rozenberg
103k1891195
103k1891195
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
|
show 4 more comments
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
Sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:48
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
@yuanming luo I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 14 at 8:52
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
Thanks, I think I can apply the same tech for the negative side.
$endgroup$
– yuanming luo
Jan 14 at 8:55
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
@yuanming luo Yes, of course!
$endgroup$
– Michael Rozenberg
Jan 14 at 8:56
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
$begingroup$
How can I know the result found by AM-GM is the lower bound at the first quadrant?
$endgroup$
– yuanming luo
Jan 16 at 0:13
|
show 4 more comments
$begingroup$
The equation $$y={x^2+1over x+1}=ato x^2-ax-(a-1)=0$$for $xne -1$ has a real root if and only if $$Deltage0 iff a^2+4a-4ge 0iff age 2sqrt 2-2text{ or }ale -2sqrt 2-2$$in that case, the root is $ne -1$ automatically (if not, it leads to the impossible non-equality $a-1=a+1$), Therefore the range simply is $$(-infty,-2sqrt 2-2] bigcup [2sqrt 2-2,infty)$$
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
The equation $$y={x^2+1over x+1}=ato x^2-ax-(a-1)=0$$for $xne -1$ has a real root if and only if $$Deltage0 iff a^2+4a-4ge 0iff age 2sqrt 2-2text{ or }ale -2sqrt 2-2$$in that case, the root is $ne -1$ automatically (if not, it leads to the impossible non-equality $a-1=a+1$), Therefore the range simply is $$(-infty,-2sqrt 2-2] bigcup [2sqrt 2-2,infty)$$
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
The equation $$y={x^2+1over x+1}=ato x^2-ax-(a-1)=0$$for $xne -1$ has a real root if and only if $$Deltage0 iff a^2+4a-4ge 0iff age 2sqrt 2-2text{ or }ale -2sqrt 2-2$$in that case, the root is $ne -1$ automatically (if not, it leads to the impossible non-equality $a-1=a+1$), Therefore the range simply is $$(-infty,-2sqrt 2-2] bigcup [2sqrt 2-2,infty)$$
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
The equation $$y={x^2+1over x+1}=ato x^2-ax-(a-1)=0$$for $xne -1$ has a real root if and only if $$Deltage0 iff a^2+4a-4ge 0iff age 2sqrt 2-2text{ or }ale -2sqrt 2-2$$in that case, the root is $ne -1$ automatically (if not, it leads to the impossible non-equality $a-1=a+1$), Therefore the range simply is $$(-infty,-2sqrt 2-2] bigcup [2sqrt 2-2,infty)$$
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 9:31
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
$begingroup$
Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=frac{x^2+1}{x+1}$
$$Rightarrow yx + y = x^2 +1 Rightarrow -x^2 + yx +(y-1)=0 Rightarrow x = frac{-y pmsqrt{y^2 + 4(y-1)}}{-2} Rightarrow frac{-y pm sqrt{(y-(-2-sqrt5))(y - (sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = frac{-sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
$endgroup$
add a comment |
$begingroup$
Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=frac{x^2+1}{x+1}$
$$Rightarrow yx + y = x^2 +1 Rightarrow -x^2 + yx +(y-1)=0 Rightarrow x = frac{-y pmsqrt{y^2 + 4(y-1)}}{-2} Rightarrow frac{-y pm sqrt{(y-(-2-sqrt5))(y - (sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = frac{-sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
$endgroup$
add a comment |
$begingroup$
Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=frac{x^2+1}{x+1}$
$$Rightarrow yx + y = x^2 +1 Rightarrow -x^2 + yx +(y-1)=0 Rightarrow x = frac{-y pmsqrt{y^2 + 4(y-1)}}{-2} Rightarrow frac{-y pm sqrt{(y-(-2-sqrt5))(y - (sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = frac{-sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
$endgroup$
Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=frac{x^2+1}{x+1}$
$$Rightarrow yx + y = x^2 +1 Rightarrow -x^2 + yx +(y-1)=0 Rightarrow x = frac{-y pmsqrt{y^2 + 4(y-1)}}{-2} Rightarrow frac{-y pm sqrt{(y-(-2-sqrt5))(y - (sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = frac{-sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
answered Jan 14 at 10:47
Abhas Kumar SinhaAbhas Kumar Sinha
18715
18715
add a comment |
add a comment |
$begingroup$
Since $$lim_{xto-infty} frac{x^2+1}{x+1} = -infty,$$
the minimum does not exist.
answered Jan 14 at 8:46
5xum5xum
90.8k394161
$endgroup$
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
add a comment |
$begingroup$
Since $$lim_{xto-infty} frac{x^2+1}{x+1} = -infty,$$
the minimum does not exist.
answered Jan 14 at 8:46
5xum5xum
90.8k394161
$endgroup$
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
add a comment |
$begingroup$
Since $$lim_{xto-infty} frac{x^2+1}{x+1} = -infty,$$
the minimum does not exist.
answered Jan 14 at 8:46
5xum5xum
90.8k394161
$endgroup$
Since $$lim_{xto-infty} frac{x^2+1}{x+1} = -infty,$$
the minimum does not exist.
answered Jan 14 at 8:46
5xum5xum
90.8k394161
answered Jan 14 at 8:46
5xum5xum
90.8k394161
answered Jan 14 at 8:46
5xum5xum
90.8k394161
answered Jan 14 at 8:46
5xum5xum
90.8k394161
90.8k394161
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
add a comment |
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
$begingroup$
sorry, I use the misunderstood word. It's a local maximum.
$endgroup$
– yuanming luo
Jan 14 at 8:47
add a comment |
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math.stackexchange.com/questions/174905/…
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– lab bhattacharjee
Jan 14 at 12:40