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Exercise:
Consider a function $g:\mathbb{R}\to \mathbb{R}$ defined to be$\left( {{C}_{g}} \right)_{{}}^{{}}$the graph of $g\left( x \right)=1-x{{e}^{-x}}$
1) a) Calculate $\underset{x\to \pm \infty }{\mathop{\lim }}\,g\left( x \right)$ and deduce the asymptote to $\left( {{C}_{g}} \right)$
b) Calculate $g'\left( x \right)$ and setup the table of variations of $g$
c) Calculate $g\left( -1 \right)\,\,\And g\left( 0 \right)\,\,$and plot $\left( {{C}_{g}} \right)$
2) Let $f:\mathbb{R}\to \mathbb{R}$ be a function defined to be $\left( {{C}_{f}} \right)$ the graph $f\left( x \right)=x+\left( x+1 \right){{e}^{-x}}$
a) Calculate $\underset{x\to \pm \infty }{\mathop{\lim }}\,f\left( x \right)$
b) Show that, the line $\left( d \right):y=x$ is an asymptote to $\left( {{C}_{f}} \right)$
c) Study the relative position of $\left( {{C}_{f}} \right)\And \left( d \right)$
d) Show that , $f'=g$ and set up the table of variation of $f$
e) Calculate $f\left( -2 \right)\And f\left( 0 \right)$ and plot $\left( {{C}_{f}} \right)$
3) a) Calculate the area bounded by $\left( {{C}_{f}} \right)$ ,$\left( d' \right):x=0$ and $\left( d'' \right):x=\ln 3$
b) show that, $f$ admits inverse function $h$ to be determined
c) Determine the domain of definition of $h$ and plot its graph $\left( {{C}_{h}} \right)$
d) Write the equation of tangent $\left( T \right)$ to $\left( {{C}_{h}} \right)$ at the point $\left( 1,0 \right)$
Solution:
1) a) $\underset{x\to \infty }{\mathop{\lim }}\,g\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,1-x{{e}^{-x}}=1-\underset{x\to \infty }{\mathop{\lim }}\,x{{e}^{-x}}=1-\underset{x\to \infty }{\mathop{\lim }}\,\frac{x}{{{e}^{x}}}=1-\frac{1}{\infty }=1-0=1$
similarly we get $\underset{x\to -\infty }{\mathop{\lim }}\,g\left( x \right)=1$ hence $y=1$ is H.A
b) $g'\left( x \right)=-\left( x{{e}^{-x}} \right)'=-\left( {{e}^{-x}}+\left( -{{e}^{-x}} \right)x \right)=-{{e}^{-x}}+x{{e}^{-x}}={{e}^{-x}}\left( x-1 \right)$
but ${{e}^{-x}}>0$ hence $g'\left( x \right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1$ and $g\left( 1 \right)=1-{{e}^{-1}}\approx 0.6321$
$\begin{align}
& \underline{\,\,\,\,\,\left. x\,\,\,\,\, \right|-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty }\, \\
& \underline{\left. \,g'(x)\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& \left. \,\,\,g(x)\, \right|\,{{\,}^{1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\searrow \,\,\,\,\,\,{{\,}_{0.6321}}\,\,\,\,\,\,\,\,\,\,\nearrow \,\,\,\,\,\,\,\,\,\,\,\,{{\,}^{1}} \\
\end{align}$
c) $g\left( -1 \right)=1+e\,\,\,\,,g\left( 0 \right)=1$
2) a) $\underset{x\to \pm \infty }{\mathop{\lim }}\,f\left( x \right)=\underset{x\to \pm \infty }{\mathop{\lim }}\,x+\left( x+1 \right){{e}^{-x}}=\pm \infty +\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{x+1}{{{e}^{x}}}=\pm \infty $ then O.A
b) $\underset{x\to \infty }{\mathop{\lim }}\,\left( f\left( x \right)-x \right)=\underset{x\to \infty }{\mathop{\lim }}\,\frac{x+1}{{{e}^{x}}}=0$ hence $y=x$ is an asymptote
c) if $\left( d \right)\cap \left( {{C}_{f}} \right)=\left\{ p \right\}\Leftrightarrow {{y}_{d}}=f\left( x \right)\Leftrightarrow x+1=0\Leftrightarrow x=-1$
if $\left( d \right)$ above $\left( {{C}_{f}} \right)$ then $x>-1$ below otherwise
d). $f'\left( x \right)=1+{{e}^{-x}}-{{e}^{-x}}\left( x+1 \right)=1+{{e}^{-x}}\left( 1-x-1 \right)=1-x{{e}^{-x}}=g\left( x \right)$
we have $g\left( x \right)$ above x-axis hence $g\left( x \right)>0\Leftrightarrow f'>0\,\,\forall x\in \mathbb{R}$
$\begin{align}
& \underline{\,\,\,\,\,\left. x\,\,\,\,\, \right|-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty }\, \\
& \underline{\left. \,f'(x)\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
& \left. \,\,\,f(x)\, \right|\,{{\,}_{y=x}}\,\,\,\,\,\,\,\,\,\,\nearrow \,\,\,\,\,\,\,\,\,\,\,\,{{\,}^{y=x}} \\
\end{align}$
e). $f\left( -2 \right)=-2+\left( -2+1 \right){{e}^{2}}=-2-{{e}^{2}}$ and
$f\left( 0 \right)=1$
3) $A=\int_{0}^{\ln 3}{f\left( x \right)dx}=\int_{0}^{\ln 3}{\left( x+\left( x+1 \right){{e}^{-x}} \right)dx}=\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{\ln 3}+\int_{0}^{\ln 3}{\left( x+1 \right){{e}^{-x}}dx}$
$=\frac{{{\ln }^{2}}3}{2}+\int_{0}^{\ln 3}{x{{e}^{-x}}dx+\int_{0}^{\ln 3}{{{e}^{-x}}dx}=\ln 3+\int_{0}^{\ln 3}{x{{e}^{-x}}dx-\left[ {{e}^{^{-x}}} \right]_{0}^{\ln 3}}}$
$=\ln 3-\left( {{e}^{-\ln 3}}-{{e}^{0}} \right)+\int_{0}^{\ln 3}{x{{e}^{-x}}dx}=\ln 3+\frac{1}{3}-1-\left( x{{e}^{-x}} \right)_{0}^{\ln 3}+\int_{0}^{\ln 3}{{{e}^{-x}}dx}$
$=\frac{1}{6}\left( 8+3{{\ln }^{2}}3-\ln 9 \right)\approx 1.5706\,\,uni{{t}^{2}}$
Since $f$ is strictly increasing hence monotonic and continuous hence
${{f}^{-1}}$ exists which is $h\left( x \right)={{f}^{-1}}\left( x \right)$ where the domain of $h={{R}_{f}}$
a) . we know that $h\left( x \right)={{f}^{-1}}\left( x \right)\Leftrightarrow h'\left( x \right)=\frac{1}{f'\left( {{f}^{-1}}\left( x \right) \right)}=\frac{1}{g\left( {{f}^{-1}}\left( x \right) \right)}$
So the slope of $\left( T \right)$ is ${{\left. h'\left( x \right) \right|}_{x=1}}=\frac{1}{g\left( {{f}^{-1}}\left( 1 \right) \right)}=\frac{1}{g\left( 0 \right)}=\frac{1}{1}$ So $\left( T \right):y=x-1$
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