10

Homework 10.pdf

1. (20 points) If \(f : (0, \infty) \to (0, \infty)\) is differentiable at \(a \in (0, \infty)\), then evaluate \(\lim_{x \to a} \left( \frac{f(x)}{f(a)} \right)^{\frac{1}{\ln(x) - \ln(a)}}\)

   We have \(\begin{aligned}\lim_{x\to a}\left(\frac{f(x)}{f(a)}\right)^{\frac{1}{\ln(x)-\ln(a)}}=\lim_{x\to a}e^{\ln\left(\frac{f(x)}{f(a)}\right)^{\frac{1}{\ln(x)-\ln(a)}}}\end{aligned}=\lim_{x\to a}e^{\frac{1}{\ln(x)-\ln(a)}\ln\left(\frac{f(x)}{f(a)}\right)}=\lim_{x\to a}e^{\frac{\ln\left(\frac{f(x)}{f(a)}\right)}{\ln(x)-\ln(a)}}=e^{\lim_{x\to a}\frac{\ln\left(\frac{f(x)}{f(a)}\right)}{\ln(x)-\ln(a)}}\)

   Since \(f\) is differentiable on \((0,\infty)\) and \(a\in (0,\infty)\), then it is also continuous on \((0,\infty)\).

   And since \(\ln\) is differentiable and continuous on \((0,\infty)\), then we can apply the L'Hopitals' Rule

   \(\begin{aligned}e^{\lim_{x\to a}\frac{\ln\left(\frac{f(x)}{f(a)}\right)}{\ln(x)-\ln(a)}}=e^{\lim_{x\to a}\frac{\frac{f\left(a\right)}{f\left(x\right)}\cdot\frac{f^{\prime}\left(x\right)}{f\left(a\right)}}{\frac{1}{x}}}=e^{\frac{\frac{f^{\prime}\left(a\right)}{f\left(a\right)}}{\frac{1}{a}}}=e^{\frac{af^{\prime}\left(a\right)}{f\left(a\right)}}\end{aligned}\)​

   Thus \(\lim_{x\to a}\left(\frac{f(x)}{f(a)}\right)^{\frac{1}{\ln(x)-\ln(a)}}=e^{\frac{af^{\prime}\left(a\right)}{f\left(a\right)}}\)

2. (20 points) Prove that \(x - \frac{x^3}{3!} < \sin x\)​ for all \(x > 0\).

   Since \(x - \frac{x^3}{3!} < \sin x\iff\sin x-x+\frac{x^3}{6}>0\), then consider \(f(x)=\sin x-x+\frac{x^3}{6}\)

   We need to prove \(f(x)>0,\forall x>0\)​

   \(f^{\prime}(x)=\cos x-1+\frac{x^2}{2}\) and \(f''(x)=-\sin x+x\)

   Since we know \(\sin x<x\), \(\forall x\in[0,\frac{\pi}{2}]\), then \(f''(x)>0,\forall x\in [0,\frac{\pi}{2}]\)

   Since when \(x>\frac{\pi}{2}\), \(\sin x\leq 1\), then \(f''(x)>0,\forall x\in (0,\infty)\)

   Then \(f'(x)\) is strictly monotonic increasing

   Since \(f'(0)=1-1+0=0\), then \(\forall x\in (0,\infty)\), \(f'(x)>0\)

   Thus \(f(x)\) is strictly monotonic increasing.

   Since \(f(0)=0-0+0=0\), then \(f(x)>0,\forall x>0\)

   Thus \(x - \frac{x^3}{3!} < \sin x\) for all \(x > 0\).

3. (60 points) For a given function \(f\) determine:

   • (a) \(f(x) = \frac{x^2}{1 - x^2}\)

   • i) Domain

     Since \(1-x^2\neq 0\Rightarrow x\neq \pm1\), then \(x\in(-\infty,-1)\cup(-1,1)\cup\left(1,\infty\right)\)​ - ii) \(x\)-intercept and \(y\)-intercept

     Let \(f(x)=0\Rightarrow\frac{x^2}{1-x^2}=0\Rightarrow x^2=0\Rightarrow x=0\)​

     Let \(x=0\Rightarrow f\left(x\right)=0\)​

     Thus \(x\)-intercept and \(y\)-intercept are all \((0,0)\)​ - iii) Symmetry

     Since \(f(x)=f(-x)\), then it is symmetric of \(y\)-axis - iv) Asymptotes

     \(\lim_{x\to\infty}\frac{x^2}{1-x^2}=\lim_{x\to\infty}\frac{1}{\frac{1}{x^2}-1}=\frac{1}{\lim_{x\to\infty}\frac{1}{x^2}-1}=-1=\lim_{x\to-\infty}=\frac{x^2}{1-x^2}\) \(\Rightarrow y=-1\) is horizontal asymptote

     Since \(\lim_{x\to-1^{+}}f\left(x\right)=-\infty=\lim_{x\to1^{-}}f\left(x\right)\) and \(\lim_{x\to-1^{-}}f\left(x\right)=\infty=\lim_{x\to1^{+}}f\left(x\right)\)​

     Then \(x=\pm 1\) is vertical asymptote - v) Increasing and decreasing intervals

     \(f(x)=\frac{x^2}{1-x^2}\Rightarrow f^{\prime}\left(x\right)=\frac{2x\left(1-x^2\right)-x^2\left(-2x\right)}{\left(1-x^2\right)^2}=\frac{2x-2x^3+2x^3}{\left(1-x^2\right)^2}=\frac{2x}{\left(1-x^2\right)^2}=0\Rightarrow x=0\)​

     Then

     	\(f'(x)\)​	\(f(x)\)​
     \((-\infty, 0)\)​	\(-\)​	Strictly monotonic decreasing
     \((0,\infty)\)​	\(+\)​	Strictly monotonic increasing
     - vi) Local maximum and minimum

     Since the form above, then \(x=0\) is local minimum - vii) Concavity intervals. Inflection points.

     \(f''\left(x\right)=\frac{2\cdot\left(1-x^2\right)^2-2x\cdot2\left(1-x^2\right)\left(-2x\right)}{\left(1-x^2\right)^4}=\frac{2\cdot\left(1-x^2\right)-2x\cdot2\left(-2x\right)}{\left(1-x^2\right)^3}=\frac{2-2x^2+8x^2}{\left(1-x^2\right)^3}=\frac{6x^2+2}{\left(1-x^2\right)^3}\)​

     \(f''(x)>0\Rightarrow\) \(1-x^2>0\Rightarrow x\in (-1,1)\) concave up

     \(f''(x)<0\Rightarrow\) \(1-x^2<0\Rightarrow x\in (-\infty,-1)\cup(1,\infty)\) concave down

     \(x=-1,1\) are inflection points - viii) Sketch of the graph

     ​ - ix) Global maximum and minimum

     There is no global maximum and global minimum - (b) \(f(x) = x \ln(x)\)

   • i) Domain \(x\in (0,\infty)\)​

   • ii) \(x\)-intercept and \(y\)-intercept

     \(f(x)=0\Rightarrow x=1\) which is \(x\)-intercept

     \(x=0\) impossible - iii) Symmetry

     No - iv) Asymptotes

     \(\lim_{x\to\infty}x\ln\left(x\right)=\infty\)

     \(\lim_{x\to0^{+}}x\ln\left(x\right)=\lim_{x\to0^{+}}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to0^{+}}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to0^{+}}-x=0\)

     There is no asymptotes - v) Increasing and decreasing intervals

     \(f^{\prime}(x)=\ln(x)+x\frac{1}{x}=\ln x+1=0\Rightarrow x=\frac{1}{e}\)​

     	\(f'(x)\)​	\(f(x)\)​
     \((0,\frac{1}{e})\)​	\(-\)​	Strictly monotonic decreasing
     \((\frac{1}{e},\infty)\)​	\(+\)​	Strictly monotonic increasing
     - vi) Local maximum and minimum

     From above, we know \(x=\frac1e\) is local minimum, there is no local maximum - vii) Concavity intervals. Inflection points.

     \(f''(x)=(\ln x+1)^{\prime}=\frac{1}{x}\)​

     There is no inflection points and \(x>0\), thus the function is always concave up - viii) Sketch of the graph

     - ix) Global maximum and minimum

     Thus the global maximum doesn't exist, the global minimum is \(-\frac1e\)​ - (c) \(f(x) = \arctan(3x) - \arctan(x)\)/

   • i) Domain \(\R\)

   • ii) \(x\)-intercept and \(y\)-intercept

     \(f(x)=0\Rightarrow\arctan(3x)-\arctan(x)=0\Rightarrow\arctan(3x)=\arctan(x)\Rightarrow3x=x\Rightarrow x=0\)​

     which is \(x\)-\(y\) intercept - iii) Symmetry

     \(f(x)=-f(-x)\) which is symmetric about zero point - iv) Asymptotes

     \(\lim_{x\to\infty}\arctan(3x)-\arctan(x)=\lim_{x\to\infty}\arctan(3x)-\lim_{x\to\infty}\arctan(x)=0\) which is horizontal asymptote - v) Increasing and decreasing intervals

     \(f^{\prime}(x)=\frac{3}{1+9x^2}-\frac{1}{1+x^2}=0\Rightarrow1+9x^2=3+3x^2\Rightarrow6x^2=2\Rightarrow x=\pm\frac{\sqrt3}{3}\)​

     Thus

     	\(f'(x)\)​	\(f(x)\)​
     \((-\infty,-\frac{\sqrt3}{3})\)​	\(-\)​	Strictly monotonic decreasing
     \((-\frac{\sqrt3}{3},\frac{\sqrt3}{3})\)​	\(+\)​	Strictly monotonic increasing
     \((\frac{\sqrt3}{3},\infty)\)​	\(-\)​	Strictly monotonic decreasing
     - vi) Local maximum and minimum

     From above \(x=-\frac{\sqrt3}{3}\) is local minimum and \(x=\frac{\sqrt3}{3}\) is local maximum - vii) Concavity intervals. Inflection points.

     \(f''(x)=-\frac{54x}{\left(1+9x^2\right)^2}+\frac{2x}{\left(1+x^2\right)^2}=0\Rightarrow54x+54x^5+108x^3=2x+162x^5+36x^3\)

     Then \(108x^5-72x^3-52x=0\Rightarrow27x^5-18x^3-13x=0\Rightarrow x=0,\pm\frac{\sqrt{3+4\sqrt3}}{3}\)

     	\(f''(x)\)​	\(f(x)\)​
     \((-\infty,-\frac{\sqrt{3+4\sqrt3}}{3})\)​	\(-\)​	concave down
     \((-\frac{\sqrt{3+4\sqrt3}}{3},0)\)​	\(+\)​	concave up
     \((0,\frac{\sqrt{3+4\sqrt3}}{3})\)​	\(-\)​	concave down
     \((\frac{\sqrt{3+4\sqrt3}}{3},\infty)\)​	\(+\)​	concave up

     Thus the reflection points are \(x=0,\pm\frac{\sqrt{3+4\sqrt3}}{3}\)​ - viii) Sketch of the graph

     ​ - ix) Global maximum and minimum

     From above, we know the global minimum is \(f(-\frac{\sqrt3}{3})=-\frac{\pi}{6}\) and the global maximum is \(f(\frac{\sqrt3}{3})=\frac{\pi}{6}\)

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