113. Range for a function.
by Virgil Nicula, Sep 12, 2010, 2:54 AM
PP1. Prove that 
![$x\in \left[ 0,\frac {\pi}{2} \right]$](//latex.artofproblemsolving.com/8/a/d/8ad43ede177553cee0460d41973edb7b83c0b293.png)
have 
where the number 
is the unique solution of the equation 
Proof. Let
, . Thus, 
, i.e. 
is strict decreasing and 
. Hence exists
and is uniquelly
so that 
and 
, i.e. 
on 
and 
on 
. In conclusion 
on ![$[0,a]$](//latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
and 
on ![$\left[a,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/2/4/9/249076ac21cfc16f7f39517eef08db3b4c5e4348.png)
, i.e.
on ![$\left[ 0,\frac {\pi}{2} \right]$](//latex.artofproblemsolving.com/2/6/9/2699cb7d0cdbcc327713edcb58420183d2bb1a9a.png)
. Since 
obtain 
, i.e. 
for any .
PP2. Without using calculus, find the maximum and minimum values on the curve
.
Proof. Observe that
and 
, where the function 
is odd and
![$f(x)\in (-\infty ,-4]\cup [4,\infty )$](//latex.artofproblemsolving.com/7/e/4/7e4040ee42952b8d537837e0c32f3dd6fd6f69ee.png)
for any 
because 
. Therefore, ![$\frac 1y\in (-\infty ,-1]\cup [7,\infty )$](//latex.artofproblemsolving.com/6/c/f/6cfd7304533ee56cca216799775c8ae527d2d9b4.png)
, i.e.
![$y\in [-1,0)\cup\left (0,\frac 17\right]=\left[-1,\frac 17\right]^*$](//latex.artofproblemsolving.com/0/8/1/081e86a2502b29c4513d622c18102357a2e06d79.png)
. For 
obtain 
. In conclusion, ![$y\in \left[-1,\frac 17\right]$](//latex.artofproblemsolving.com/d/b/7/db73a0d301137813782a87d1d429b5a795ca0273.png)
.
PP3. Find the range of the function
.
Method 1. Denote
. Observe that 
and for 
,
where
. The function 
is odd and for 
. In conclusion, for any we have 
![$g(x)\in (-\infty ,-2]\cup [2,\infty )\implies$](//latex.artofproblemsolving.com/1/b/b/1bbadbb2924df58380dbdf08057f766380e4bdd6.png)
![$f(x)\in \left[\frac 12,\frac 32\right]$](//latex.artofproblemsolving.com/2/3/2/232fd33008bfb5ae2f5ab563bccbf354bf441bcb.png)
. In conclusion, for any 
we have ![$f(x)\in \left[\frac 12,\frac 32\right]\cup\{0\}$](//latex.artofproblemsolving.com/f/4/3/f4344d9b07066de79014fb653974f82f94f58368.png)
, i.e. ![$y\in \left[\frac 12,\frac 32\right]$](//latex.artofproblemsolving.com/b/f/d/bfd15a8ad9d5b9cf773f767a17a251b3f5a368c5.png)
.
Method 2. Denote
, i.e. 
. For 
have 
. Let 
. I'"ll find 
for which
so that 
, i.e. the discriminant 
![$y\in\left[\frac 12,\frac 32\right]$](//latex.artofproblemsolving.com/9/5/1/951ef8a546de78049bb1bebe88b59a9e74818da6.png)
.
PP4.[/url] Prove that![$(\forall )\ x\in\mathbb R^*\ ,\ x\left[(x^2+1)^{\frac{1}{x}}-e^{2\arctan{x}}\right]< 0\ .$](//latex.artofproblemsolving.com/3/c/1/3c15bf2c73eb583385348a96214f615687a40ed1.png)
Proof. Define the relation over
, i.e. 
and 
have same signature. Denote the function 
, where
. Observe that 
and 
, i.e. 
. Thus, for 
, i.e. (increasing) and for 
, i.e. (decreasing). Thus, and 
for any 
![$(\forall )\ x\in\mathbb R^*\ ,\ \left[(x^2+1)^{\frac 1x}-e^{2\arctan x}\right]\ \mathrm{.s.s.}$](//latex.artofproblemsolving.com/1/0/d/10d50b148afe91e39a9e8ad1d77354451692aa84.png)
![$x\left[(x^2+1)^{\frac{1}{x}}-e^{2\arctan{x}}\right]< 0$](//latex.artofproblemsolving.com/d/9/a/d9a9ef59b0874fb2d585ec5ee9e4b02cc77744b4.png)
, i.e. 
.
![$x\in \left[ 0,\frac {\pi}{2} \right]$](http://latex.artofproblemsolving.com/8/a/d/8ad43ede177553cee0460d41973edb7b83c0b293.png)
have 
where the number 
is the unique solution of the equation 
Proof. Let
, . Thus, 
, i.e. 
is strict decreasing and 
. Hence existsand is uniquelly
so that 
and 
, i.e. 
on 
and 
on 
. In conclusion 
on ![$[0,a]$](http://latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
and 
on ![$\left[a,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/2/4/9/249076ac21cfc16f7f39517eef08db3b4c5e4348.png)
, i.e.
on ![$\left[ 0,\frac {\pi}{2} \right]$](http://latex.artofproblemsolving.com/2/6/9/2699cb7d0cdbcc327713edcb58420183d2bb1a9a.png)
. Since 
obtain 
, i.e. 
for any .PP2. Without using calculus, find the maximum and minimum values on the curve
.Proof. Observe that
and 
, where the function 
is odd and![$f(x)\in (-\infty ,-4]\cup [4,\infty )$](http://latex.artofproblemsolving.com/7/e/4/7e4040ee42952b8d537837e0c32f3dd6fd6f69ee.png)
for any 
because 
. Therefore, ![$\frac 1y\in (-\infty ,-1]\cup [7,\infty )$](http://latex.artofproblemsolving.com/6/c/f/6cfd7304533ee56cca216799775c8ae527d2d9b4.png)
, i.e.![$y\in [-1,0)\cup\left (0,\frac 17\right]=\left[-1,\frac 17\right]^*$](http://latex.artofproblemsolving.com/0/8/1/081e86a2502b29c4513d622c18102357a2e06d79.png)
. For 
obtain 
. In conclusion, ![$y\in \left[-1,\frac 17\right]$](http://latex.artofproblemsolving.com/d/b/7/db73a0d301137813782a87d1d429b5a795ca0273.png)
.PP3. Find the range of the function
.Method 1. Denote
. Observe that 
and for 
,where
. The function 
is odd and for 
. In conclusion, for any we have 
![$g(x)\in (-\infty ,-2]\cup [2,\infty )\implies$](http://latex.artofproblemsolving.com/1/b/b/1bbadbb2924df58380dbdf08057f766380e4bdd6.png)
![$f(x)\in \left[\frac 12,\frac 32\right]$](http://latex.artofproblemsolving.com/2/3/2/232fd33008bfb5ae2f5ab563bccbf354bf441bcb.png)
. In conclusion, for any 
we have ![$f(x)\in \left[\frac 12,\frac 32\right]\cup\{0\}$](http://latex.artofproblemsolving.com/f/4/3/f4344d9b07066de79014fb653974f82f94f58368.png)
, i.e. ![$y\in \left[\frac 12,\frac 32\right]$](http://latex.artofproblemsolving.com/b/f/d/bfd15a8ad9d5b9cf773f767a17a251b3f5a368c5.png)
.Method 2. Denote
, i.e. 
. For 
have 
. Let 
. I'"ll find 
for which
so that 
, i.e. the discriminant 
![$y\in\left[\frac 12,\frac 32\right]$](http://latex.artofproblemsolving.com/9/5/1/951ef8a546de78049bb1bebe88b59a9e74818da6.png)
.PP4.[/url] Prove that
![$(\forall )\ x\in\mathbb R^*\ ,\ x\left[(x^2+1)^{\frac{1}{x}}-e^{2\arctan{x}}\right]< 0\ .$](http://latex.artofproblemsolving.com/3/c/1/3c15bf2c73eb583385348a96214f615687a40ed1.png)
Proof. Define the relation over
, i.e. 
and 
have same signature. Denote the function 
, where
. Observe that 
and 
, i.e. 
. Thus, for 
, i.e. (increasing) and for 
, i.e. (decreasing). Thus, and 
for any 
![$(\forall )\ x\in\mathbb R^*\ ,\ \left[(x^2+1)^{\frac 1x}-e^{2\arctan x}\right]\ \mathrm{.s.s.}$](http://latex.artofproblemsolving.com/1/0/d/10d50b148afe91e39a9e8ad1d77354451692aa84.png)
![$x\left[(x^2+1)^{\frac{1}{x}}-e^{2\arctan{x}}\right]< 0$](http://latex.artofproblemsolving.com/d/9/a/d9a9ef59b0874fb2d585ec5ee9e4b02cc77744b4.png)
, i.e. 
.This post has been edited 19 times. Last edited by Virgil Nicula, Nov 30, 2015, 2:33 PM
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