Problem source

The function $\mathrm{f}$ is defined for $x<2$ by 
\[
\mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|.
\]
Find the maximum and minimum points and the points of inflection of the graph of $\mathrm{f}$ and sketch this graph. Is $\mathrm{f}$ continuous everywhere? Is $\mathrm{f}$ differentiable everywhere? 
Find the inverse of the function $\mathrm{f}$, i.e. expressions for $\mathrm{f}^{-1}(x),$ defined in the various appropriate intervals.

Solution source

\[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \]

Therefore the absolute value terms will change behaviour at $x = -1, 0, 1$. Then

\begin{align*}
f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) &  x \leq -1 \\
2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\
-2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\
2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\
 &= \begin{cases} x^2-4x-1 &  x \leq -1 \\
 3x^2+1& -1 < x \leq 0 \\
-5x^2+1& 0 < x \leq 1 \\
x^2-4x-1 & 1 < x\end{cases} \\
\\
f'(x)  &= \begin{cases} 2x-4 &  x <-1 \\
 6x & -1 < x  < 0 \\
-10x & 0 < x < 1 \\
2x-4 & 1 < x\end{cases} \\
\end{align*}

Therefore $f'(x) = 0 \Rightarrow x = 0, 2$ and so we should check all the turning points.

Therefore the minimum is $x = 2, y = -5$, maximum is $x = -2, y = 11$ (assuming the range is actually $|x| < 2$. There is a point of inflection at $x = 0, y = 1$.

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-2.5};
    \def\xu{2.5};
    \def\yl{-5};
    \def\yu{11};
    
    % 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=-2:-1, samples=100] 
            plot (\x, {(\x)^2-4*\x-1});
        \draw[thick, blue, smooth, domain=-1:0, samples=100] 
            plot (\x, {3*(\x)^2+1});
        \draw[thick, blue, smooth, domain=0:1, samples=100] 
            plot (\x, {-5*(\x^2)+1});
        \draw[thick, blue, smooth, domain=1:2, samples=100] 
            plot (\x, {(\x)^2-4*\x-1});

        
    \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}

$f$ is continuous everywhere as a sum of continuous functions. $f$ is not differentiable at $x = -1, 1$



Suppose 
\begin{align*}
&&y &=x^2-4x-1 \\
&&&= (x-2)^2 -5 \\
\Rightarrow &&x &= 2\pm \sqrt{y+5} \\
\\
&& y &= 3x^2+1 \\
\Rightarrow && x &= \pm \sqrt{\frac{y-1}{3}} \\
\\
&& y &= -5x^2+1 \\
\Rightarrow && x &=\pm \sqrt{\frac{1-y}{5}} \\
\\
\Rightarrow && f^{-1}(y) &= \begin{cases} 2 - \sqrt{y+5} & y > 4 \\
 -\sqrt{\frac{y-1}{3}} & 1 < y < 4 \\
\sqrt{\frac{1-y}{5}} & -4 < y < 1 \\
2 + \sqrt{y+5} & y < -4
\end{cases}
\end{align*}