214. A range of some functions.
by Virgil Nicula, Jan 25, 2011, 11:36 AM
Proposed problem 1. Find the range of the function 
, where 
.
Proof. Observe that
, where 
and 
.
Thus
and 
the function 
is decreasing and 
.
Proposed problem 2. Let
be two real numbers such that 
. Find the range of 
.
Proof. Observe that
. If 
, then 
. If 
, then 
, where 
. Thus, 
and
. Therefore, 
. Hence 
and 
.
Proposed problem 3. Find the range of the function
, where 
.
Proof.![$u(x)=\sin x+\cos x=t\in\left[1,\sqrt 2\right]$](//latex.artofproblemsolving.com/b/0/7/b07d66566e47afc2eb4bbb2f9175db2956e1c7fd.png)
, where 
.
Proposed problem 4. Find the minimum value of
, where 
.
Proof. Denote
. I""ll find the minimum value of 
, where 
. Using AM-GM inequality obtain that
![$\left\{\begin{array}{ccccc}
9t^2+\frac 6t+\frac 6t\ge 3\cdot\sqrt [3]{9\cdot 6\cdot 6}=9\cdot\sqrt [3]{12} & \mathrm{with\ equality\ iff} & 9t^2=\frac 6t & \iff & t=\sqrt [3]{\frac 23}\\\\
\frac {4}{t^2}+6t+6t\ge 3\cdot\sqrt [3]{4\cdot 6\cdot 6}=6\cdot\sqrt [3]{18} & \mathrm{with\ equality\ iff} & \frac {4}{t^2}=6t & \iff & t=\sqrt [3]{\frac 23}\end{array}\right\|$](//latex.artofproblemsolving.com/6/e/c/6ec8a3121946de46952f467c41f96650df9976b7.png)
.
Hence![$9\tan^2x+4\cot^2x+12 \tan x+12 \cot x\ge 9\cdot\sqrt [3]{12}+6\cdot\sqrt [3]{18}=3\sqrt[3]6\cdot\left(3\sqrt [3]2+2\sqrt [3]3\right)$](//latex.artofproblemsolving.com/d/6/c/d6c3bc25ea01ba3a442c6a7097a8a366af300daa.png)
with equality iff ![$\tan x=\sqrt [3]{\frac 23}$](//latex.artofproblemsolving.com/0/d/8/0d8ac5b3c094999a4b977637f544f331ec472faa.png)
.
An easy extension. Find the minimum value of
, where 
and .
PP5. Ascertain the range of the function
.
Proof. is positive and even, 
, where 
and ![$h(t)=\sqrt{1+2t\sqrt {1-t^2}}+\sqrt {1-2t\sqrt{1-x^2}}\ ,\ t\in [0,1]$](//latex.artofproblemsolving.com/e/2/9/e2950efde1378544a726a43b4b532acd691179a2.png)
.
Thus,
, i.e. the function 
and the function
. In conclusion, ![$\mathrm{Im}(f)=f\left(\left[-1,1\right]\right)=\mathrm{Im}(h)=h\left(\left[0,1\right]\right)=\left[\sqrt 2,2\right]$](//latex.artofproblemsolving.com/9/8/e/98ebe13bb094e1468b58b041f174b415462305f1.png)
and for any
the equation 
has the solution 
, where ![$\mathrm{card}\ S_{\lambda}(f)=\left\{\begin{array}{ccc}
0 & \mathrm {if} & \lambda\not\in \left[\sqrt 2,2\right]\\\
2 & \mathrm {if} & \lambda =\sqrt 2\\\
3 & \mathrm{if} & \lambda =2\\\
4 & \mathrm{if} & \lambda\in \left(\sqrt 2,2\right)\end{array}\right\|$](//latex.artofproblemsolving.com/a/5/c/a5cc32f39fe1812cf32a152daaf522c4a244e9dc.png)
.
The graph
of the function show so : http://www.artofproblemsolving.com/Forum/download/file.php?mode=view&id=38351&
PP6. Find the range of the function
.
Proof. Observe that
and 
. Consider ![$f:(0,1]\rightarrow \mathbb R$](//latex.artofproblemsolving.com/8/9/5/895636006d350e0fd5a79706ea86dd0c15e8bedf.png)
, where 
. Observe that 
, where ![$t=\phi (x)=x+\frac 1x\ ,\ \phi : (0,1]\rightarrow [2,\infty )$](//latex.artofproblemsolving.com/6/7/9/6795c2c5d4b45d94e426a1e99b948d17bf1dfec6.png)
is bijective decreasing (
) and 
is bijective increasing (
). In conclusion, 
, where ![$I=(0,1]$](//latex.artofproblemsolving.com/c/3/3/c33d2b70b5dc6fca5d1417f1d7b59362f630d493.png)
.
Therefore,![$\left\{\begin{array}{ccc}
f\left(\frac 1x\right)=f(x)\ ,\ (\forall )\ x\in (0,1] & \implies & f\left(\mathbb R^*_+\right)=f(I)=\left[\frac 13,1\right)\\\\
f(-x)=\frac {1}{f(x)}\ ,\ (\forall )\ x>0 & \implies & f\left(\mathbb R^*_-\right)=(1,3]\end{array}\right\|$](//latex.artofproblemsolving.com/2/4/5/245b20c0444f5566ea4871c4a8724e1bab3e66bd.png)
. In conclusion, 
![$\{1\}\cup\left[\frac 13,1\right)\cup (1,3]\implies$](//latex.artofproblemsolving.com/4/f/4/4f42b5404e74f5c380771c8fd1e5d7f5fe88978f.png)
![$\boxed{f(\mathbb R)=\left[\frac 13,3\right]}$](//latex.artofproblemsolving.com/0/1/1/01111471eb6b3be7ea9f2b69506b78704db88e43.png)
.
, where 
.Proof. Observe that
, where 
and 
.Thus
and 
the function 
is decreasing and 
.Proposed problem 2. Let
be two real numbers such that 
. Find the range of 
.Proof. Observe that
. If 
, then 
. If 
, then 
, where 
. Thus, 
and
. Therefore, 
. Hence 
and 
.Proposed problem 3. Find the range of the function
, where 
.Proof.
![$u(x)=\sin x+\cos x=t\in\left[1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/b/0/7/b07d66566e47afc2eb4bbb2f9175db2956e1c7fd.png)
, where 
.Proposed problem 4. Find the minimum value of
, where 
.Proof. Denote
. I""ll find the minimum value of 
, where 
. Using AM-GM inequality obtain that![$\left\{\begin{array}{ccccc}
9t^2+\frac 6t+\frac 6t\ge 3\cdot\sqrt [3]{9\cdot 6\cdot 6}=9\cdot\sqrt [3]{12} & \mathrm{with\ equality\ iff} & 9t^2=\frac 6t & \iff & t=\sqrt [3]{\frac 23}\\\\
\frac {4}{t^2}+6t+6t\ge 3\cdot\sqrt [3]{4\cdot 6\cdot 6}=6\cdot\sqrt [3]{18} & \mathrm{with\ equality\ iff} & \frac {4}{t^2}=6t & \iff & t=\sqrt [3]{\frac 23}\end{array}\right\|$](http://latex.artofproblemsolving.com/6/e/c/6ec8a3121946de46952f467c41f96650df9976b7.png)
.Hence
![$9\tan^2x+4\cot^2x+12 \tan x+12 \cot x\ge 9\cdot\sqrt [3]{12}+6\cdot\sqrt [3]{18}=3\sqrt[3]6\cdot\left(3\sqrt [3]2+2\sqrt [3]3\right)$](http://latex.artofproblemsolving.com/d/6/c/d6c3bc25ea01ba3a442c6a7097a8a366af300daa.png)
with equality iff ![$\tan x=\sqrt [3]{\frac 23}$](http://latex.artofproblemsolving.com/0/d/8/0d8ac5b3c094999a4b977637f544f331ec472faa.png)
. An easy extension. Find the minimum value of
, where 
and .PP5. Ascertain the range of the function
.Proof.
is positive and even, 
, where 
and ![$h(t)=\sqrt{1+2t\sqrt {1-t^2}}+\sqrt {1-2t\sqrt{1-x^2}}\ ,\ t\in [0,1]$](http://latex.artofproblemsolving.com/e/2/9/e2950efde1378544a726a43b4b532acd691179a2.png)
.Thus,
, i.e. the function 
and the function
. In conclusion, ![$\mathrm{Im}(f)=f\left(\left[-1,1\right]\right)=\mathrm{Im}(h)=h\left(\left[0,1\right]\right)=\left[\sqrt 2,2\right]$](http://latex.artofproblemsolving.com/9/8/e/98ebe13bb094e1468b58b041f174b415462305f1.png)
and for any
the equation 
has the solution 
, where ![$\mathrm{card}\ S_{\lambda}(f)=\left\{\begin{array}{ccc}
0 & \mathrm {if} & \lambda\not\in \left[\sqrt 2,2\right]\\\
2 & \mathrm {if} & \lambda =\sqrt 2\\\
3 & \mathrm{if} & \lambda =2\\\
4 & \mathrm{if} & \lambda\in \left(\sqrt 2,2\right)\end{array}\right\|$](http://latex.artofproblemsolving.com/a/5/c/a5cc32f39fe1812cf32a152daaf522c4a244e9dc.png)
.The graph
of the function show so : http://www.artofproblemsolving.com/Forum/download/file.php?mode=view&id=38351&PP6. Find the range of the function
.Proof. Observe that
and 
. Consider ![$f:(0,1]\rightarrow \mathbb R$](http://latex.artofproblemsolving.com/8/9/5/895636006d350e0fd5a79706ea86dd0c15e8bedf.png)
, where 
. Observe that 
, where ![$t=\phi (x)=x+\frac 1x\ ,\ \phi : (0,1]\rightarrow [2,\infty )$](http://latex.artofproblemsolving.com/6/7/9/6795c2c5d4b45d94e426a1e99b948d17bf1dfec6.png)
is bijective decreasing (
) and 
is bijective increasing (
). In conclusion, 
, where ![$I=(0,1]$](http://latex.artofproblemsolving.com/c/3/3/c33d2b70b5dc6fca5d1417f1d7b59362f630d493.png)
.Therefore,
![$\left\{\begin{array}{ccc}
f\left(\frac 1x\right)=f(x)\ ,\ (\forall )\ x\in (0,1] & \implies & f\left(\mathbb R^*_+\right)=f(I)=\left[\frac 13,1\right)\\\\
f(-x)=\frac {1}{f(x)}\ ,\ (\forall )\ x>0 & \implies & f\left(\mathbb R^*_-\right)=(1,3]\end{array}\right\|$](http://latex.artofproblemsolving.com/2/4/5/245b20c0444f5566ea4871c4a8724e1bab3e66bd.png)
. In conclusion, 
![$\{1\}\cup\left[\frac 13,1\right)\cup (1,3]\implies$](http://latex.artofproblemsolving.com/4/f/4/4f42b5404e74f5c380771c8fd1e5d7f5fe88978f.png)
![$\boxed{f(\mathbb R)=\left[\frac 13,3\right]}$](http://latex.artofproblemsolving.com/0/1/1/01111471eb6b3be7ea9f2b69506b78704db88e43.png)
. This post has been edited 45 times. Last edited by Virgil Nicula, Nov 22, 2015, 4:00 PM
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