Grade 10: Exercise (4.6) - Solution

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A real valued function is one-to-one if every horizontal line intersects the graph of the function at most one point.

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1. Determine whether each of the following function is a one-to-one function or not. If it is not one-to-one, explain why not.

(a) It is a one to one function.

(b) It is not a one to one function because some horizontal lines intersect the graph of the function more than one point.

(c) It is a one to one function.

(d) It is not a one to one function because some horizontal lines intersect the graph of the function more than one point.

(c) It is a one to one function.

(c) It is a one to one function.

2. Draw the graph of the each given function and determine whether each is a one-to-one function or not.
   
   (a) $f(x)=3 x+2$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \ldots & -2 & -1 & 0 & 1 & 2 & \ldots \\ \hline f(x) & \cdots & -4 & -1 & 2 & 5 & 8 & \ldots \\ \hline \end{array}$
   
   

   It is a one to one function.

   (b) $f(x)=x-3$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \ldots & -2 & -1 & 0 & 1 & 2 & . \\ \hline f(x) & \cdots & -5 & -4 & -3 & 2 & 1 & \cdots \\ \hline \end{array}$
   
   

   It is a one to one function.

   (c) $f(x)=4 x^{2}$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \cdots & -2 & -1 & 0 & 1 & 2 & \cdots \\ \hline f(x) & \cdots & 16 & 4 & 0 & 4 & 16 & \cdots \\ \hline \end{array}$
   
   

   It is not a one to one function.

   (d) $f(x)=2|x|$
   
   $\begin{array}{|c||c|c|c|c|c|c|c|} \hline x & \cdots- & -2 & -1 & 0 & 1 & 2 & \cdots \\ \hline f(x) & \cdots & 4 & 2 & 0 & 2 & 4 & \cdots \\ \hline \end{array}$
   
   

   It is not a one to one function.

   (e) $f(x)=\dfrac{2 x+3}{x+2}$
   
   $\begin{aligned} f(x) &=\dfrac{2 x+3}{x+2} \\\\ &=\dfrac{2 x+4-1}{(x+2)} \\\\ &=\dfrac{-1+2(x+2)}{(x+2)} \\\\ &=\dfrac{-1}{x+2}+2\\\\ &\text { horizontal arymptote: } y=2 \\\\ &\text { vertical asymptote : } x=-2 \\\\ &x=0, y=\frac{3}{2} \\\\ &y \text { -intercept }=\left(0, \frac{3}{2}\right) \\\\ &y=0, x=-\frac{3}{2} \\\\ &x \text { -intercept }=\left(-\frac{3}{2}, 0\right) \end{aligned}$
   
   

   It is a one to one function.

   (f) $f(x)=4 x^{2}\quad (0 \le x \le 4)$
   
   $\begin{array}{|c||c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 0 & 4 & 16 & 36 & 64 \\ \hline \end{array}$
   
   

   It is a one to one function.

   (g) $f(x)=\sqrt{x}\quad (x \ge 0)$
   
   $\begin{array}{|c||c|c|c|c|c|c|} \hline x & 0 & 1 & 4 & 9 & 25 & \ldots \\ \hline f(x) & 0 & 1 & 2 & 3 & 5 & \cdots \\ \hline \end{array}$
   
   

   It is a one to one function.

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