Problem source

Give a sketch of  the curve  $ \;\displaystyle y= \frac1 {1+x^2}\;$,  for $x\ge0$. 
Find the equation of the line that intersects the curve at $x=0$ and is tangent to the curve at some point with $x>0\,$. Prove that there are no further intersections between the line and the curve. Draw the line on your sketch.
By considering the area under the curve for $0\le x\le1$, show that $\pi>3\,$. Show also, by considering the volume formed by rotating the curve about the $y$ axis, that $\ln 2 >2/3\,$.

[\textbf{Note}: $\displaystyle \int_0^ 1 \frac1 {1+x^2}\, \d x = \frac\pi 4\,.\;$]

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/(1+(#1)^2)};
    \def\xl{-0.2};
    \def\xu{5};
    \def\yl{-0.2};
    \def\yu{1.2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=0:\xu, samples=100] 
            plot (\x, {\functionf(\x)});

        \draw[red] (\xl, {1-0.5 * \xl}) -- (\xu, {1-0.5 * \xu});


    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\begin{align*}
&& y &= (1+ x^2)^{-1} \\
\Rightarrow && y' &= -2x(1+x^2)^{-2} \\
\text{eqn of tangent}:&& \frac{y - (1+t^2)^{-1}}{x-t} &= -2t(1+t^2)^{-2} \\
\text{passes thru }(0,1): && \frac{1-(1+t^2)^{-1}}{-t} &= -2t(1+t^2)^{-2} \\
\Rightarrow && (1+t^2)^2-(1+t^2) &= 2t^2 \\
\Rightarrow && t^4-t^2 &= 0 \\
\Rightarrow && t &= 0, \pm 1 \\
\Rightarrow && \frac{y - \frac12}{x - 1} &= -\frac12 \\
&& y &=1 -\tfrac12 x 
\end{align*}

There can be no further intersections since the equation is equivalent to the cubic $(1-\frac12 x)(1+x^2) = 1$ and we have already found $3$ roots.

\begin{align*}
&& A &= \int_0^1 \frac{1}{1 + x^2} = \frac{\pi}{4}  \\
&& A &> \frac12 \cdot 1 \cdot (1 + \tfrac12) = \frac34 \\
\Rightarrow && \pi &> 3 \\
\\
&& V &=\pi \int_{\frac12}^1 x^2 \d y \\
&&&= \pi \int_{\frac12}^1 \left ( \frac{1}{y}-1 \right) \d y \\
&&&= \pi \left [\ln y \right]_{1/2}^1-\frac12 \\
&&&= \pi \ln 2 - \frac{\pi}{2} \\
&& V &> \frac13 \pi 1^2 \frac{1}{2}  \\
&&&= \frac{\pi}{6} \\
\Rightarrow && \ln 2 &> \frac{2}{3}
\end{align*}