Problem source

The function $\mathrm{Min}$ is defined as
\[
\mathrm{Min}(a, b) = \begin{cases}
a & \text{if } a \leq b \\
b & \text{if } a > b
\end{cases}
\]
\begin{questionparts}
\item Sketch the graph $y = \mathrm{Min}(x^2, 2x)$.
\item Solve the equation $2\mathrm{Min}(x^2, 2x) = 5x - 3$.
\item Solve the equation $\mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) = mx$ in the cases $m = 2$ and $m = 6$.
\item Show that $(1, -3)$ is a local maximum point on the curve $y = 2\mathrm{Min}(x^2, x^3) - 5x$ and find the other three local maxima and minima on this curve.
Sketch the curve.
\end{questionparts}

Solution source

\begin{questionparts}

\item \begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-5};
    \def\yu{10};
    
    % 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, red, dashed, domain=-5:5, samples=100] 
            plot (\x, {2*\x});
        \draw[thick, red, dashed, domain=-5:5, samples=100] 
            plot (\x, {\x*\x});
        \draw[thick, blue, smooth, domain=-5:5, samples=100] 
            plot (\x, {min(\x*\x, 2*\x)});  
        % \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        % \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}

\item $2 \textrm{Min}(x^2,2x) = 5x-3$ tells us either $2x^2 = 5x-3 \Rightarrow 2x^2 - 5x +3 = 0 \Rightarrow (2x-3)(x-1) \Rightarrow x = 1, \frac32$ and $0 \leq x \leq 2$ or $4x = 5x-3 \Rightarrow x= 3$ and $x < 0$ or $2 > x$, therefore our solutions are $x = 1, \frac32, 3$

\item We have different cases based on $x$ vs $-2, 0, 2$, ie

Case $x \leq -2$:
\begin{align*}
&& mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\
&&&= 2x + x^3
\end{align*}

So $2x = 2x + x^3 \Rightarrow x^3 = 0$, but $x \leq -2$ so no solutions.

or $6x = 2x + x^3 \Rightarrow 0 = x(x^2-4) \Rightarrow x = 0, 2, -2$ so $x = -2$.

Case $-2  < x \leq 0$:
\begin{align*}
&& mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\
&&&= 2x + 4x
\end{align*}

So $2x = 2x + 4x$ ie $x = 0$ which is valid.

Or $6x = 2x + 4x$ ie valid for all values in $-2 \leq x \leq 0$

Case $0  < x \leq 2$:
\begin{align*}
&& mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\
&&&= x^2 + x^3
\end{align*}

So $2x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-2) = x(x-1)(x+2)$ so $x = 0, 1, -2$, but the range means $x = 0$ or $x = 1$

Or $6x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-6) = x(x-2)(x+3)$ so $x = 0, 2, -3$, but the range means $x = 0$ or $x = 2$

Case $2  \leq  x $:
\begin{align*}
&& mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\
&&&=2x + 4x
\end{align*}

So $2x = 2x + 4x \Rightarrow x = 0$ so no solutions.

Or $6x = 2x + 4x$ so a range of solutions.


Therefore the final solutions for $m = 2$ are $x = 0, x = 1$ and for $m = 2$ are $x \in [-2,0] \cup [2, \infty)$

\item $\mathrm{Min}(x^2, x^3)$ switches when $x = 1$, so we must consider both limits:

\begin{align*}
&& \frac{\d y}{\d x}\vert_{x > 1} &= 4x - 5 \\
\\
&& \frac{\d y}{\d x}\vert_{x < 1} &= 6x^2 - 5 \\
\end{align*}

so when $x = 1$ the sign of the derivative changes from positive to negative, hence a local maximum.

The other local maxima and minima will be when $x = \frac54$ or $x = \pm \sqrt{5/6}$


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-5};
    \def\yu{10};
    
    % 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, red, dashed, domain=-5:5, samples=100] 
            plot (\x, {2*\x*\x-5*\x});
        \draw[thick, red, dashed, domain=-5:5, samples=100] 
            plot (\x, {2*\x*\x*\x-5*\x});
        \draw[thick, blue, smooth, domain=-5:5, samples=100] 
            plot (\x, {2*min(\x*\x, \x*\x*\x) - 5*\x});  
        % \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
        % \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
        % \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
    \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}

\end{questionparts}