Problem source

Using the series
\[
\e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,,
\]
show that $\e>\frac83$. 
Show that   $n!>2^n$ for $n\ge4$ and hence show that
$\e<\frac {67}{24}$.

Show that the curve with equation 
\[
y= 3\e^{2x} +14 \ln (\tfrac43-x)\,,
\qquad
{x<\tfrac43}
\]
has a minimum turning point between $x=\frac12$ and $x=1$ and 
give a  sketch to show the shape of the curve.

Solution source

\begin{align*}
&& e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\
&&&> 1 + 1+ \frac12 + \frac16 \\
&&&= \frac{12+3+1}{6} = \frac83
\end{align*}

$4! = 24 > 16 = 2^4$, notice that $n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n$.

\begin{align*}
&& e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\
&&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\
&&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\
&&&= \frac83 + \frac1{8} \\
&&&= \frac{67}{24}
\end{align*}

\begin{align*}
&& y &= 3e^{2x} +14 \ln(\tfrac43-x) \\
&& y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\
&& y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\
&&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\
&& y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\
&&&= 6e^2 - 42 = 6(e^2-7) \\
&&&> 6(\tfrac{64}{9} - 7) > 0
\end{align*}

Therefore $y'$ changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){3*exp(2*(#1))+14*ln(4/3-(#1))};
    \def\xl{-2};
    \def\xu{2};
    \def\yl{-5};
    \def\yu{20};
    
    % 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=\xl:{4/3-0.01}, samples=201] 
            plot ({\x}, {\functionf(\x)});

        
        \draw[thin, red, dashed, domain=\xl:\xu, samples=100] 
            plot ({\x},{3*exp(2*\x)});
        \draw[thin, red, dashed, domain=\xl:{4/3-0.01}, samples=100] 
            plot ({\x},{14*ln(4/3-\x)});
        % \draw[thick, blue, smooth, domain=\yl:-1, samples=100] 
        %     plot ({-(sqrt(\x*\x-1)-0.25)}, {\x});
        
    \end{scope}

    % \node[below] at ({pi/2}, 0) {$\frac{\pi}2$};
    \node[below] at (0.5,0) {$\frac12$};
    \node[below] at (1,0) {$1$};
    
    % 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}