Problem source

The function $\mathrm{f}$ is defined by 
\[
\mathrm{f}(x)=\frac{\left(x-a\right)\left(x-b\right)}{\left(x-c\right)\left(x-d\right)}\qquad\left(x\neq c,\ x\neq d\right),
\]
where $a,b,c$ and $d$ are real and distinct, and $a+b\neq c+d$.
Show that 
\[
\frac{x\mathrm{f}'(x)}{\mathrm{f}(x)}=\left(1-\frac{a}{x}\right)^{-1}+\left(1-\frac{b}{x}\right)^{-1}-\left(1-\frac{c}{x}\right)^{-1}-\left(1-\frac{d}{x}\right)^{-1},
\]
$(x\neq0,x\neq a,x\neq b)$ and deduce that when $\left|x\right|$ is much larger than each of $\left|a\right|,\left|b\right|,\left|c\right|$ and $\left|d\right|,$ the gradient of $\mathrm{f}(x)$ has the same
sign as $(a+b-c-d).$ 
It is given that there is a real value of real value of $x$ for which $\mathrm{f}(x)$ takes the real value $z$ if and only if 
\[
[\left(c-d\right)^{2}z+\left(a-c\right)\left(b-d\right)+\left(a-d\right)\left(b-c\right)]^{2}\geqslant4\left(a-c\right)\left(b-d\right)\left(a-d\right)\left(b-c\right).
\]
Describe briefly a method by which this result could be proved, but do not attempt to prove it.  
Given that $a < b$ and $a < c < d$, make sketches of the graph of $\mathrm{f}$ in the four distinct cases which arise, indicating the cases for which the range of $\mathrm{f}$ is not the whole of $\mathbb{R}.$

Solution source

Notice that $\ln f(x) = \ln (x - a) + \ln (x-b) - \ln (x-c) - \ln (x-d)$ therefore:

\begin{align*}
\frac{\d}{\d x}: && \frac{f'(x)}{f(x)} &= (x-a)^{-1}+(x-b)^{-1}-(x-c)^{-1} - (x-d)^{-1} \\
&&&= \frac{1}{x} \left ( (1-\frac{a}{x})^{-1}+(1-\frac{b}{x})^{-1}-(1-\frac{c}{x})^{-1} - (1-\frac{d}{x})^{-1}\right)
\end{align*}

Multiplying by $x$ gives the desired result.

When $|x|$ is very large then:

\begin{align*}
\frac{x f'(x)}{f(x)} &\approx 1 + \frac{a}{x} + o(\frac{1}{x^2})+ 1 + \frac{b}{x} + o(\frac{1}{x^2})-(1 + \frac{c}{x} + o(\frac{1}{x^2}))-(1 + \frac{d}{x} + o(\frac{1}{x^2})) \\
&= \frac{a+b-c-d}{x} + o(x^{-2})
\end{align*}

Dividing by $x$ we obtain $\frac{f'(x)}{f(x)} \approx \frac{a+b-c-d}{x^2} + o(x^{-3})$ if $|x|$ is sufficiently large this will be dominated by the $\frac{a+b-c-d}{x^2}$ term which will have the same sign as $(a+b-c-d)$.  When $|x|$ is very large all of the brackets will have the same sign, and therefore $f(x)$ will be positive, and so $f'(x)$ must have the same sign as $a+b-c-d$.

To prove this result, we could set $f(x) = k$ and rearrange to form a quadratic in $x$. We could then check the discriminant is non-zero.

Case 1: $a < c < d < b$ and $a+b > c+d \Rightarrow$ not all values reached and approx asymtope from below on the right and above on the left.

\begin{center}
\begin{tikzpicture}[scale=1.2]
  % Define the function
  \def\functionf(#1){1/50*(#1+1)*(#1-3)/(#1*(#1-1))}
  
  \draw[->] (-3.5,0) -- (5.5,0) node[right] {$x$};
  \draw[->] (0,-4.5) -- (0,4.5) node[above] {$f(x)$};

  \begin{scope}
    \clip (-3.5,-4.5) rectangle (5.5,4.5);
    
    % Vertical asymptotes
    \draw[red, dashed] (0,-4.5) -- (0,4.5);
    \draw[red, dashed] (1,-4.5) -- (1,4.5);
    
    % Horizontal asymptote
    \draw[green!60!black, dashed] (-3.5,1) -- (5.5,1);
    
    % Plot the function in segments to avoid asymptotes
    % Left of x=0
    \draw[blue, thick, smooth] plot[domain=-3:-0.1, samples=100] (\x, {50*\functionf(\x)});
    
    % Between x=0 and x=1
    \draw[blue, thick, smooth] plot[domain=0.1:0.9, samples=50] (\x, {10*\functionf(\x)});
    
    % Right of x=1
    \draw[blue, thick, smooth] plot[domain=1.1:5, samples=100] (\x, {50*\functionf(\x)});
  \end{scope}
  
  
  % Mark x-intercepts
  \filldraw[black] (-1,0) circle (2pt);
  \filldraw[black] (3,0) circle (2pt);
\end{tikzpicture}
\end{center}

Case 2: $a < c < d < b$ and $a + b < c + d \Rightarrow$ not all values hit, but approach from above on the right and below on the left

\begin{center}
\begin{tikzpicture}[scale=1.2]
  % Define the function
  \def\functionf(#1){1/50*(#1+1)*(#1-1.5)/(#1*(#1-1))}
  
  \draw[->] (-3.5,0) -- (5.5,0) node[right] {$x$};
  \draw[->] (0,-4.5) -- (0,4.5) node[above] {$f(x)$};

  \begin{scope}
    \clip (-3.5,-4.5) rectangle (5.5,4.5);
    
    % Vertical asymptotes
    \draw[red, dashed] (0,-4.5) -- (0,4.5);
    \draw[red, dashed] (1,-4.5) -- (1,4.5);
    
    % Horizontal asymptote
    \draw[green!60!black, dashed] (-3.5,1) -- (5.5,1);
    
    % Plot the function in segments to avoid asymptotes
    % Left of x=0
    \draw[blue, thick, smooth] plot[domain=-3:-0.1, samples=100] (\x, {50*\functionf(\x)});
    
    % Between x=0 and x=1
    \draw[blue, thick, smooth] plot[domain=0.1:0.9, samples=50] (\x, {10*\functionf(\x)});
    
    % Right of x=1
    \draw[blue, thick, smooth] plot[domain=1.1:5, samples=100] (\x, {50*\functionf(\x)});
  \end{scope}
  
  
  % Mark x-intercepts
  \filldraw[black] (-1,0) circle (2pt);
  \filldraw[black] (1.5,0) circle (2pt);
\end{tikzpicture}
\end{center}

Case 2: $a < c < b < d \Rightarrow$ all values hit

\begin{center}
\begin{tikzpicture}[scale=1.2]
  % Define the function
  \def\functionf(#1){1/50*(#1+1)*(#1-0.5)/(#1*(#1-1))}
  
  \draw[->] (-3.5,0) -- (5.5,0) node[right] {$x$};
  \draw[->] (0,-4.5) -- (0,4.5) node[above] {$f(x)$};

  \begin{scope}
    \clip (-3.5,-4.5) rectangle (5.5,4.5);
    
    % Vertical asymptotes
    \draw[red, dashed] (0,-4.5) -- (0,4.5);
    \draw[red, dashed] (1,-4.5) -- (1,4.5);
    
    % Horizontal asymptote
    \draw[green!60!black, dashed] (-3.5,1) -- (5.5,1);
    
    % Plot the function in segments to avoid asymptotes
    % Left of x=0
    \draw[blue, thick, smooth] plot[domain=-3:-0.1, samples=100] (\x, {50*\functionf(\x)});
    
    % Between x=0 and x=1
    \draw[blue, thick, smooth] plot[domain=0.1:0.9, samples=50] (\x, {10*\functionf(\x)});
    
    % Right of x=1
    \draw[blue, thick, smooth] plot[domain=1.1:5, samples=100] (\x, {50*\functionf(\x)});
  \end{scope}
  
  
  % Mark x-intercepts
  \filldraw[black] (-1,0) circle (2pt);
  \filldraw[black] (0.5,0) circle (2pt);
\end{tikzpicture}
\end{center}

Case 3: $a < b < c < d \Rightarrow $ not all values hit

\begin{center}
\begin{tikzpicture}[scale=1.2]
  % Define the function
  \def\functionf(#1){1/50*(#1+1)*(#1+0.5)/(#1*(#1-1))}
  
  \draw[->] (-3.5,0) -- (5.5,0) node[right] {$x$};
  \draw[->] (0,-4.5) -- (0,4.5) node[above] {$f(x)$};

  \begin{scope}
    \clip (-3.5,-4.5) rectangle (5.5,4.5);
    
    % Vertical asymptotes
    \draw[red, dashed] (0,-4.5) -- (0,4.5);
    \draw[red, dashed] (1,-4.5) -- (1,4.5);
    
    % Horizontal asymptote
    \draw[green!60!black, dashed] (-3.5,1) -- (5.5,1);
    
    % Plot the function in segments to avoid asymptotes
    % Left of x=0
    \draw[blue, thick, smooth] plot[domain=-3:-0.1, samples=100] (\x, {50*\functionf(\x)});
    
    % Between x=0 and x=1
    \draw[blue, thick, smooth] plot[domain=0.1:0.9, samples=50] (\x, {10*\functionf(\x)});
    
    % Right of x=1
    \draw[blue, thick, smooth] plot[domain=1.1:5, samples=100] (\x, {50*\functionf(\x)});
  \end{scope}
  
  
  % Mark x-intercepts
  \filldraw[black] (-1,0) circle (2pt);
  \filldraw[black] (-0.5,0) circle (2pt);
\end{tikzpicture}
\end{center}