9

Homework9.pdf

December 21, 2024

1. Let \(f:[a,b]\rightarrow[a,b]\) be differentiable and \(|f^{\prime}(x)|\neq1\) for all \(x\in[a,b]\) . Then \(f\) has a unique fixed point in \([a,b]\) .

   Proof

   We need to prove there exists unique \(c\in[a,b]\) such that \(f(c)=c\), then \(g(x)=f(x)-x\) exists only one root.

   Existence: Since \(g(a)=f(a)-a\geq 0\) and \(g(b)=f(b)-b\leq 0\)​

   By intermedia value theorem, there exists \(c\in[a,b]\) such that \(g(c)=f(c)-c=0\), then \(f(c)=c\)​

   Uniqueness: Suppose there exists two different such \(c\)​

   Then \(g(c_1)=f(c_1)-c_1=0=g(c_2)=f(c_2)-c_2\).

   Then by mean value theorem, there exists \(\tilde{c}\in\left(c_1,c_2\right):f^{\prime}\left(\tilde{c{}}\right)=\frac{f\left(c_2\right)-f\left(c_1\right)}{c_2-c_1}=\frac{c_2-c_1}{c_2-c_1}=1\)​

   Contradiction!

   Thus there is a unique fixed point in \([a,b]\) .

2. Find the number of (distinct) real roots of the equation \(e^{2x}+\cos x+x=0\) .

   Consider \(f(x)=e^{2x}+\cos x+x\), then \(f^{\prime}(x)=e^{2x}\cdot2-\sin x+1\)​

   Since \(1-\sin x\geq 0\) and \(2e^{2x}>0\), then \(f'(x)>0\), then \(f(x)\) is strictly increasing

   Then the equation only has at most one root

   Since \(f(-100)<0\) and \(f(100)>0\), then by Bolzano Theorem there exists one root of \(f(x)\)

   Thus the number of distinct real roots of the equation is one

3. Let \(f,g:I\rightarrow\mathbb{R}\) differentiable on \(I\) , and \(a\in I\) . Show that if \(f^{\prime}(x)>g^{\prime}(x)\) for all \(x\in I\) , and \(f(a)=g(a)\), then \(f\!\left(x\right)>g\!\left(x\right)\) for all \(x>a\) and \(f\!\left(x\right)<g\!\left(x\right)\) for all \(x<a\) .

   Consider \(h(x)=f(x)-g(x)\). Since \(f'(x)>g'(x)\), then \(f'(x)-g'(x)>0\Rightarrow h'(x)>0\), then \(h\) is strictly increasing.

   Also since \(f(a)=g(a)\), then \(f(a)-g(a)=h(a)=0\)​

   Then we need to prove \(f\!\left(x\right)>g\!\left(x\right)\) for all \(x>a\) and \(f\!\left(x\right)<g\!\left(x\right)\) for all \(x<a\)

   Which means we need to prove \(h(x)>0\) for all \(x>a\) and \(h(x)<0\) for all \(x<a\)​

   Since \(h(a)=0\) and \(h\) is strictly increasing, then \(h(x)>0\) for all \(x>a\) and \(h(x)<0\) for all \(x<a\) automatically.

   Then \(f\!\left(x\right)>g\!\left(x\right)\) for all \(x>a\) and \(f\!\left(x\right)<g\!\left(x\right)\) for all \(x<a\) .

4. Prove the following inequality \(2\sin x+\tan x>3x,{\mathrm{~for~all~}}x\in\left(0,{\frac{\pi}{2}}\right)\)

   Consider \(f(x)=2\sin x+\tan x-3x\) for all \(x\in\left(0,{\frac{\pi}{2}}\right)\). Then \(f^{\prime}(x)=2\cos x+\frac{1}{\cos^2x}-3\).

   Then consider \(f^{\prime}(x)=0\Rightarrow2\cos x+\frac{1}{\cos^2x}-3=0\Rightarrow2\cos^3x-3\cos^2x+1=0\Rightarrow\cos x=1,-0.5\)​

   Since \(x\in\left(0,{\frac{\pi}{2}}\right)\), then there isn't such \(x\) such that \(f'(x)=0\).

   Since \(f^{\prime}(\frac{\pi}{4})=\sqrt2+2-3>0\), thus the function is strictly increasing when \(x\in\left(0,{\frac{\pi}{2}}\right)\)​

   Since \(f(0)=2+1-3=0\), then \(f(x)>0\) for all \(x\in\left(0,{\frac{\pi}{2}}\right)\)​

   Thus \(2\sin x+\tan x>3x,{\mathrm{~for~all~}}x\in\left(0,{\frac{\pi}{2}}\right)\)

5. Determine for the following functions:

   • Domain

   • global, local maximum and minimum of \(f\) .

   • increasing and decreasing intervals of \(f\) .

   • An approximate graph of the functions.

   1. \(f(x)=x^{x}\)

      • Domain \(\left(0,+\infty\right)\)​

      • increasing and decreasing intervals of \(f\)

      \(f^{\prime}(x)=\left(e^{\ln x^{x}}\right)^{\prime}=\left(e^{x\ln x}\right)^{\prime}=e^{x\ln x}\cdot\left(x\ln x\right)^{\prime}=e^{x\ln x}\cdot\left(\ln x+x\frac{1}{x}\right)=e^{x\ln x}\cdot\left(\ln x+1\right)\)

      Then \(f^{\prime}(x)=0=e^{x\ln x}\cdot\left(\ln x+1\right)\Rightarrow\ln x+1=0\Rightarrow x=\frac{1}{e}\)

      Since \(f'(x)>0\) when \(x>\frac1e\) and \(f'(x)<0\) when \(0<x<\frac1e\), then \(f(x)\) is strictly increasing in \((\frac1e,+\infty)\) and strictly decreasing in \((0,\frac1e)\)​

      • global, local maximum and minimum of \(f\)

      Then from above we know \(x=\frac1e\) is a local minimum point, then \(f(\frac{1}{e})=\left(\frac{1}{e}\right)^{\frac1e}<1\) is the local minimum

      Also since \(\lim_{x\to\infty}x^{x}=\lim_{x\to\infty}e^{x\ln x}=\infty\) and \(\lim_{x\to0}x^{x}=\lim_{x\to0}e^{x\ln x}=e^0=1\), then \(f(\frac{1}{e})=\left(\frac{1}{e}\right)^{\frac1e}<1\) is the global and local minimum

      Then there is no local maximum, and the global maximum doesn't exists.

      • An approximate graph of the functions.

      ​ 2. \(f(x)=\sin(x)+\cos(x)\ x\in[0,\pi]\)

      • Domain \(x\in[0,\pi]\)

      • increasing and decreasing intervals of \(f\) .

      Since \(f(x)=\sqrt2\sin\left(x+\frac{\pi}{4}\right)\) and \(x+\frac{\pi}{4}\in [\frac{\pi}{4},\frac{5\pi}{4}]\), then \(f(x)\) is increasing in \((0,\frac{\pi}{4})\) and decreasing in \((\frac{\pi}{4},\pi)\)

      • global, local maximum and minimum of \(f\) .

      \(f(\frac{\pi}{4})=\sqrt2\sin\left(\frac{\pi}{2}\right)=\sqrt2\) which is the local and global maximum since \(f^{\prime}(\frac{\pi}{4})=\sqrt2\cos\left(x+\frac{\pi}{4}\right)=0\)

      Since \(f(0)=1\) and \(f(\pi)=-1\) , then the global and local minimum is \(-1\)

      • An approximate graph of the functions.

      ​ 3. \(g(x)=x-2\arctan(x)\)

      • Domain \(\R\)​

      • increasing and decreasing intervals of \(f\) .

      \(g'(x)=1-\frac{2}{1+x^2}\), then \(g^{\prime}(x)=0\Rightarrow1+x^2=2\Rightarrow x=\pm1\)

      Then \(x<-1\), \(g^{\prime}(x)>0\) and \(x>-1\), \(g'(x)<0\), \(x<1\), \(g'(x)<0\) and \(x>1\), \(g'(x)>0\)

      Thus \(x\in (-\infty,-1)\) is increasing interval and \(x\in (-1,1)\) is decreasing interval and \(x\in (1,\infty)\) is increasing interval

      • global, local maximum and minimum of \(f\) .

      \(f(-1)=-1-2\arctan(-1)\) is the local maximum and \(f(1)=1-2\arctan(1)\) is the local minimum

      Since \(\lim_{x\to\infty}\left(x-2\arctan(x)\right)=\infty\) and \(\lim_{x\to-\infty}\left(x-2\arctan(x)\right)=-\infty\)

      Then there is no global maximum and minimum

      • An approximate graph of the functions.

      ​ 4. \(f(x)=x^2e^{−2x^2}\)

      • Domain \(\R\)​

      • increasing and decreasing intervals of \(f\) ./

      Since \(f^{\prime}(x)=2x\cdot e^{−2x^2}+x^2\cdot e^{−2x^2}\cdot-4x=0\), we have \(2x-4x^3=0\Rightarrow x=0,\pm \frac{\sqrt2}{2}\)

      	\((-\infty,-\frac{\sqrt2}{2})\)​	\((-\frac{\sqrt2}{2},0)\)​	\((0,\frac{\sqrt2}{2})\)​	\((\frac{\sqrt2}{2},+\infty)\)​
      \(f'(x)\)​	\(>0\)​	\(<0\)​	\(>0\)​	\(<0\)​
      \(f(x)\)​	increasing	decreasing	increasing	decreasing

      • global, local maximum and minimum of \(f\) .

      Then \(f(-\frac{\sqrt2}{2})=f(-\frac{\sqrt2}{2})\) is the global and local maximum since \(\lim_{x\to\infty}=0,\lim_{x\to-\infty}=0\)

      And the global and local minimum is \(f(0)=0\)

      • An approximate graph of the functions.

      ​

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