Exercise:
Consider a function $h\left( x \right)=-{{x}^{2}}+1-\ln x$ defined on $\left( 0,\infty \right)$
1) Compute , $\underset{x\to \infty }{\mathop{\lim }}\,h\left( x \right)\,\,\,\And \,\,\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,h\left( x \right)$
2) Show that , $\frac{d}{dx}h\left( x \right)=-\left( 2x+\frac{1}{x} \right)\,\,\,\,\,\,\,\,\,\forall x\in \left( 0,\infty \right)$
3) Find the value of $h\left( 1 \right)$ and put table of variation for $h$
4) Deduce that $h\left( x \right)\ge 0\,\,\,$ when $0<x\le 1$ and $h\left( x \right)\le 0$ otherwise in $\left( 0,\infty \right)$
5) Now consider the function $f\left( x \right)=-x+\frac{\ln x}{x}$ defined on $\left( 0,\infty \right)$
i) Compute , limits of $f$ on the boundary of the interval
And deduce its asymptotes
ii) Study the relative position of $f$ with respect to $y=-x$
iii) Show that, $f'\left( x \right)=\frac{h\left( x \right)}{{{x}^{2}}}\,\,\,\forall \,x\in \left( 0,\infty \right)$
iv) Find the value of $f\left( 1 \right)$ and put the table of variation for $f$
v) Show that, $f''\left( x \right)=\frac{-3+2\ln x}{{{x}^{3}}}\,\,\,\,\,\,\forall x\in \left( 0,\infty \right)$
vi) Show that, $I=\left( {{e}^{3/2}},f\left( {{e}^{3/2}} \right) \right)$ is an inflection point for $f$
vii) Draw $f$ in the plane \(\left( O,\vec{i},\vec{j} \right)\) using ${{e}^{3/2}}\approx 4.5\,\,\,\And \,\,\,f\left( {{e}^{3/2}} \right)\approx -4.1$
Solution: we have $h:\left( 0,\infty \right)\to \mathbb{R}$ , $h\left( x \right)=-{{x}^{2}}+1-\ln x$
1) $\underset{x\to \infty }{\mathop{\lim }}\,h\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\left( -1+\frac{1}{{{x}^{2}}}-\frac{\ln x}{{{x}^{2}}} \right)=\infty \left( -1 \right)=-\infty $
But $\underset{x\to \infty }{\mathop{\lim }}\,\frac{h\left( x \right)}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{-{{x}^{2}}+1-\ln x}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\left( -x+\frac{1}{x}-\frac{\ln x}{x} \right)=-\infty $
Therefore, the asymptotic direction is parallel to $y'y$
$\underset{x\to 0}{\mathop{\lim }}\,h\left( x \right)=0+1-\ln \left( 0 \right)=\infty $ thus $x=0$ is vertical Asymptote
2) $h'\left( x \right)=-2x-\frac{1}{x}=-\left( 2x+\frac{1}{x} \right)$
3) $h'\left( x \right)=0\Leftrightarrow -2x-\frac{1}{x}=0\Leftrightarrow -2{{x}^{2}}-1<0$ hence $h$ is decreasing
$h\left( 1 \right)=-1+1-\ln 1=0$ thus $x=1$ is root for $h$
$\begin{align}
& \underline{\left. x\,\,\,\,\, \right|\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\
& \underline{\left. h'\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& \left. h\,\,\,\,\, \right|\,\,\,\,{{\,}^{+\infty }}\,\,\,\,\,\,\,\,\,\,\searrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\,}_{-\infty }} \\
\end{align}$
4) if $0<x\le 1\Leftrightarrow x>0\,\,\,\And \,\,\,x\le 1\Leftrightarrow {{x}^{2}}\le 1\Leftrightarrow -{{x}^{2}}\ge -1$
We have $x\le 1\Leftrightarrow \ln x<0\Leftrightarrow -\ln x>0\Leftrightarrow 1-\ln x>1$Thus $h\left( x \right)\ge 0$
If $x>1\Leftrightarrow -{{x}^{2}}<-1\Rightarrow -{{x}^{2}}+1<0$
But $\ln \left( x \right)>0\Rightarrow -\ln x<0$ thus $h\left( x \right)\le 0$
5) Now we take , $f:\left( 0,\infty \right)\to \mathbb{R}$ , $f\left( x \right)=-x+\frac{\ln x}{x}$
i) $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,-x+\frac{\ln x}{x}=0+\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln x}{x}=0+\frac{1}{0}=-\infty $ V.A
$\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,-x+\frac{\ln x}{x}=-\infty $ O.A
$a=\underset{x\to \infty }{\mathop{\lim }}\,\frac{f\left( x \right)}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{-x+\frac{\ln x}{x}}{x}=\underset{x\to \infty }{\mathop{\lim }}\,-1+\frac{\ln x}{{{x}^{2}}}=-1$
$b=\underset{x\to \infty }{\mathop{\lim }}\,\left( f\left( x \right)-ax \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( -x+\frac{\ln x}{x}+x \right)=0$
Therefore, $y=-x$ is Oblique asymptote
ii) Let $\left( d \right)\,:\,y=-x$ be Oblique asymptote and \(\left( c \right)\,:\,\,f\left( x \right)=-x+\frac{\ln x}{x}\)
$\left( d \right)\cap \left( c \right)=\left\{ p \right\}\Leftrightarrow \frac{\ln x}{x}=0\Leftrightarrow \ln x=\ln 1\Leftrightarrow x=1$
$\left( d \right)$ above $\left( c \right)$ if $\frac{\ln x}{x}>0$ then $x>1$
$\left( d \right)$ below $\left( c \right)$ if $\frac{\ln x}{x}<0\Leftrightarrow x<1$
iii) $f'\left( x \right)=-1+\frac{\frac{1}{x}x-\ln x}{{{x}^{2}}}=-1+\frac{1-\ln x}{{{x}^{2}}}=\frac{-{{x}^{2}}+1-\ln x}{{{x}^{2}}}=\frac{h\left( x \right)}{{{x}^{2}}}$
iv) $f\left( 1 \right)=-1+\frac{\ln 1}{1}=-1$from (4) we have ${{x}^{2}}>0$
If $0<x\le 1\Rightarrow h\left( x \right)\ge 0$ and $h\left( x \right)\le 0$ when $x\ge 1$
\[\begin{align}
& \underline{\left. x\,\,\,\, \right|0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\infty } \\
& \underline{\left. f'\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,} \\
& \left. f\,\,\, \right|\,\,\,{{\,}_{-\infty }}\,\,\,\,\,\,\,\nearrow \,\,\,\,\,\,{{\,}^{-1}}\,\,\,\,\,\,\searrow \,\,\,\,\,\,{{\,}_{-\infty }} \\
\end{align}\]
v) $f''\left( x \right)=\frac{\left( -2x-\frac{1}{x} \right){{x}^{2}}-2x\left( -{{x}^{2}}+1-\ln x \right)}{{{x}^{4}}}$
$=\frac{-2{{x}^{3}}-x+2{{x}^{3}}-2x+2x\ln x}{{{x}^{4}}}=\frac{-3x+2x\ln x}{{{x}^{4}}}=\frac{-3+2\ln x}{{{x}^{3}}}$
vi) $f''\left( x \right)=0\Leftrightarrow -3+2\ln x=0\Leftrightarrow \ln x=\frac{3}{2}\Leftrightarrow x={{e}^{3/2}}$
So $I=\left( {{e}^{3/2}},f\left( {{e}^{3/2}} \right) \right)$ is the inflection point.
vii) Figure
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