SV , SG , Grade 12 _Lebanese System Functions and Limits


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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