Mathematics Grade 12 Functions Notes Pdf

2.7 Summary (EMCFQ)

Function: a rule which uniquely associates elements of one set \(A\) with the elements of another set \(B\); each element in set \(A\) maps to only one element in set \(B\).
Functions can be one-to-one relations or many-to-one relations. A many-to-one relation associates two or more values of the independent (input) variable with a single value of the dependent (output) variable.
Vertical line test: if it is possible to draw any vertical line which crosses the graph of the relation more than once, then the relation is not a function.
Given the invertible function \(f(x)\), we determine the inverse \(f^{-1}(x)\) by:
- replacing every \(x\) with \(y\) and \(y\) with \(x\);
- making \(y\) the subject of the equation;
- expressing the new equation in function notation.
If we represent the function \(f\) and the inverse function \({f}^{-1}\) graphically, the two graphs are reflected about the line \(y=x\).
The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse.
The inverse function of a straight line is also a straight line. Vertical and horizontal lines are exceptions.
The inverse of a parabola is not a function. However, we can limit the domain of the parabola so that the inverse of the parabola is a function.
The inverse of the exponential function \(f(x) = b^{x}, (b > 0, b \ne 1)\) is the logarithmic function \(f^{-1}(x) = \log_{b}{x}\).
The "common logarithm" has a base \(\text{10}\) and can be written as \(\log_{10}{x} = \log{x}\). The \(\log\) symbol written without a base means \(\log\) base \(\text{10}\).

Logarithmic laws:

\(\log_{a}{x^{b}} = b \log_{a}{x} \qquad (x > 0)\)
\(\log_{a}{x} = \frac{\log_{b}{x}}{\log_{b}{a}} \qquad (b > 0 \text{ and } b \ne 1)\)
\(\log_{a}{xy} = \log_{a}{x} + \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\)
\(\log_{a}{\frac{x}{y}} = \log_{a}{x} - \log_{a}{y} \qquad (x > 0 \text{ and } y > 0)\)

	Straight line function	Quadratic function	Exponential function
Formula	\(y = ax + q\)	\(y = ax^{2}\)	\(y = b^{x}\)
Inverse	\(y = \frac{x}{a} - \frac{q}{a}\)	\(y = \pm \sqrt{\frac{x}{a}}\)	\(y = \log_{b}{x}\)
Inverse a function?	yes	no	yes
Graphs

End of chapter exercises

Textbook Exercise 2.12

Determine the equation of \(h\).

\begin{align*} \text{Gradient: } m &= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ &= \frac{-6 - 0}{0 - (-3)} \\ &= \frac{-6}{3} \\ &= -2\\ \therefore h(x) &= -2x - 6 \end{align*}

Find \(h^{-1}\).

\begin{align*} \text{Let } y &= -2x - 6 \\ \text{Inverse: } x &= -2y - 6 \\ x + 6 &= -2y \\ -\frac{1}{2} (x + 6) &= y \\ y &= -\frac{x}{2} - 3 \\ \therefore h^{-1}(x) &= -\frac{x}{2} - 3 \end{align*}

Draw both graphs on the same system of axes.

Calculate the coordinates of \(S\), the point of intersection of \(h\) and \(h^{-1}\).

\begin{align*} -\frac{x}{2} - 3 &= -2x - 6 \\ -x - 6 &= -4x - 12 \\ -x + 4x &= -12 + 6 \\ 3x &= -6 \\ \therefore x &= -2 \\ \text{If } x=-2, \quad y &= -2(-2) - 6 \\ y &= -2(-2) - 6 \\ &= -2 \end{align*}

This gives the point \(S(-2;-2)\)

State the property regarding the point of intersection that will always be true for a function and its inverse.

The value of the \(x\)-coordinate and the \(y\)-coordinate will always be the same since the point lies on the line \(y = x\).

Determine \(f\).

\begin{align*} \text{Let } y &= 2x + 4 \\ \text{Inverse: } x &= 2y + 4 \\ x - 4 &= 2y \\ \frac{1}{2}x - 2 &= y \\ \therefore f(x) &= \frac{1}{2}x - 2 \end{align*}

Draw \(f\) and \(f^{-1}\) on the same set of axes. Label each graph clearly.

Is \(f^{-1}\) an increasing or decreasing function? Explain your answer.

Increasing. As \(x\) increases, the function value increases. Alternative reason: gradient is positive, therefore function is increasing.

Draw the graph of \(f\) and state its domain and range.

The domain is: \(\{x: x \in \mathbb{R} \}\) and the range is: \(\{y: y \geq 0, y \in \mathbb{R} \}\).

Determine the inverse and state its domain and range.

\begin{align*} \text{Let } y &= 2x^{2} \\ \text{Inverse: } x &= 2y^{2} \\ \frac{1}{2}x &= y^{2} \\ \pm \sqrt{ \frac{1}{2}x } &= y \\ \therefore y &= \pm \sqrt{ \frac{x}{2} } \qquad (x \ge 0) \end{align*}

Domain: \(\{x: x \geq 0, x \in \mathbb{R} \}\), Range: \(\{y: y \in \mathbb{R} \}\).

Sketch the graphs of \(f\) and \(f^{-1}\) on the same system of axes.

Determine if the point \(\left( -\frac{1}{2}; 2 \right)\) lies on the graph of \(f\).

\begin{align*} f(x) & = \left( \frac{1}{4} \right)^{x} \\ \text{Substitute } \left( -\frac{1}{2}; 2 \right): \quad f\left(-\frac{1}{2}\right) & = \left( \frac{1}{4} \right)^{-\frac{1}{2}}\\ & = 4^{\frac{1}{2}} \\ & = 2 \end{align*}

Yes, the point \(\left( -\frac{1}{2}; 2 \right)\) does lie on \(f\).

Write \(f^{-1}\) in the form \(y = \ldots\)

\begin{align*} f: \quad y & = \left( \frac{1}{4} \right)^{x} \\ f^{-1}: \quad x & = \left( \frac{1}{4} \right)^{y} \\ \therefore y & = \log_{\frac{1}{4}}{x}\\ \text{or } & \\ f: \quad y & = \left( 4 \right)^{-x} \\ f^{-1}: \quad x & = \left(4 \right)^{-y} \\ -y & = \log_{4}{x}\\ \therefore y & = - \log_{4}{x} \end{align*}

\(y = \log_{\frac{1}{4}}{x}\) or \(y = - \log_{4}{x}\)

If the graphs of \(f\) and \(f^{-1}\) intersect at \(\left( \frac{1}{2}; P \right)\), determine the value of \(P\).

\(P = \frac{1}{2}\), since the point lies on the line \(y = x\).

Give the equation of the new graph, \(G\), if the graph of \(f^{-1}\) is shifted \(\text{2}\) units to the left.

Give the asymptote(s) of \(G\).

Vertical asymptote: \(x = -2\)

Write down the inverse in the form \(h^{-1}(x) = \ldots\)

\begin{align*} \text{Let: } \quad y & = 3^{x} \\ \text{Inverse: } x & = 3^{y} \\ y & = \log_{3}{x} \\ \therefore h^{-1}(x) & = \log_{3}{x} \end{align*}

State the domain and range of \(h^{-1}\).

Domain: \(\{x: x > 0, x \in \mathbb{R} \}\) and range: \(\{y: y \in \mathbb{R} \}\).

Sketch the graphs of \(h\) and \(h^{-1}\) on the same system of axes, label all intercepts.

For which values of \(x\) will \(h^{-1}(x) < 0\)?

\(0 < x < 1\)

Sketch the graphs of \(f\) and \(g\) on the same system of axes.

Determine whether or not \(f\) and \(g\) intersect at a point where \(x = -1\).

\begin{align*} f(x) & = 2^{x} \\ f(-1) & = 2^{-1} \\ & = \frac{1}{2}\\ g(x) & = x^{2} \\ g(-1) & = (-1)^{2} \\ & = 1 \\ \therefore f(-1) & \ne g(-1) \end{align*}

The graphs do not intersect at \(x = -1\).

How many solutions does the equation \(2^{x} = x^{2}\) have?

Two

Below are three graphs and six equations. Write down the equation that best matches each of the graphs.

\(y={\log}_{3}x\)
\(y=-{\log}_{3}x\)
\(y={\log}_{\frac{1}{3}}x\)
\(y={3}^{x}\)
\(y={3}^{-x}\)
\(y=-{3}^{x}\)

Graph \(\text{1}\): \(y = 3^{-x}\)

Graph \(\text{2}\): \(y = - \log _{3} x\) or \(y = \log _{\frac{1}{3}}{x}\)

Graph \(\text{3}\): \(y = -3^{x}\)

Show that \(b = 3\).

\begin{align*} y &= \log_{b}{x} \\ 2 &= \log_{b}{9} \\ \therefore 9 &= b^{2} \\ \therefore 3^2 &= b^{2} \\ \therefore b &= 3 \end{align*}

Determine the value of \(a\) if \((a;-1)\) lies on \(f\).

\begin{align*} y &= \log_{3}{x} \\ -1 &= \log_{3}{a} \\ \therefore 3^{-1} &= a \\ \therefore a &= \frac{1}{3} \end{align*}

Write down the new equation if \(f\) is shifted \(\text{2}\) units upwards.

\(y = \log_{3}{x} + 2\)

Write down the new equation if \(f\) is shifted \(\text{1}\) units to the right.

\(y = \log_{3}{(x - 1)}\)

If the rhino population in South Africa starts to decrease at a rate of \(\text{7}\%\) per annum, determine how long it will take for the current rhino population to halve in size? Give your answer to the nearest integer.

\begin{align*} A &= P \left( 1 - i \right)^{n} \\ \frac{1}{2} &= \left( 1 - \frac{7}{100}\right)^{n} \\ \text{0,5} &= \left( \text{0,93} \right)^{n} \\ \log{\text{0,5}} &= \log{\left( \text{0,93} \right)^{n}} \\ \log{\text{0,5}} &= n \log{\left( \text{0,93} \right)} \\ \therefore n &= \frac{\log{\text{0,5}} }{\log{\left( \text{0,93} \right)}} \\ &= \text{9,55} \ldots \end{align*}

It will take less than \(\text{10}\) years for the current rhino population to halve in size.

Which of the following graphs best illustrates the rhino population's decline? Motivate your answer.

Important note: the graphs above have been drawn as a continuous curve to show a trend. Rhino population numbers are discrete values and should be plotted points.

Graph C

Currently \((n = 0)\) the rhino population is \(P\). After \(\text{9,6}\) years, it will have halved, \(\frac{P}{2}\). Note: the line in the graph indicates the trend, rhino population numbers are discrete values and should be plotted points.

Determine a formula that describes this retweeting process.

\(\text{100} \qquad \text{100} \times 2 \qquad \text{100} \times 2^{2} \qquad \text{100} \times 2^{3}\)

This is a geometric sequence: \(r = 2\) and \(a = 100\).

Therefore \(T_{n} = 100 \times 2^{n-1}\).

Calculate how many retweets of the celebrity's message are sent an hour after his original tweet.

\(\text{1}\) hour \(= 60\) minutes \(= 12 \times 5\), therefore \(n = 12\)

\begin{align*} T_{n} &= 100 \times 2^{n-1} \\ T_{12} &= 100 \times 2^{11} \\ &= \text{204 800} \end{align*}

\(\text{204 800}\) retweets.

How long will it take for the total number of retweets to exceed \(\text{200}\) million?

\begin{align*} 200 \times 10^{6} &= 2 \times 10^{8} \\ S_{n} &= \frac{a(r^{n} - 1)}{r - 1} \\ \therefore \frac{100(2^{n} - 1)}{2 - 1} & > \text{2} \times 10^{\text{8}} \\ 2^{n} & > \frac{\text{2} \times 10^{\text{8}} }{100} + 1 \\ 2^{n} & > \text{2 000 001} \\ {n} & > \log_{2}{\text{2 000 001}} \qquad (\text{use definition}) \\ n & > \frac{ \log{\text{2 000 001}}}{\log2} \qquad (\text{change of base}) \\ n & > \text{20,9} \ldots \quad 5 \text{ minute periods} \end{align*}

Therefore, \(\frac{21}{12} = \text{1,75}\) hours.

Inverses (ENRICHMENT ONLY)

Textbook Exercise 2.13

Given: \(g\left(x\right)=-1+\sqrt{x}\), find the inverse of \(g\left(x\right)\) in the form \({g}^{-1}\left(x\right)=...\)

\begin{align*} g(x) & = -1 + \sqrt{x} \quad (x \ge 0) \\ \text{Let } y & = -1 + \sqrt{x} \\ \text{Inverse: } x & = -1 + \sqrt{y} \quad (y \ge 0) \\\\ \sqrt{y} & = x + 1 \qquad (x \ge -1)\\ \therefore y & = (x + 1)^{2} \\ & \\ \therefore g^{-1}(x) & = (x + 1)^{2} \qquad (x \ge -1) \end{align*}

Draw the graph of \(g^{-1}\).

Use symmetry to draw the graph of \(g\) on the same set of axes.

Is \(g^{-1}\) a function?

Yes. It passes vertical line test.

Give the domain and range of \(g^{-1}\).

Domain: \(\{x: x \ge -1, x \in \mathbb{R} \}\), Range: \(\{y: y \geq 0,y \in \mathbb{R} \}\).

Find the equation of \(f\), given that \(f\) is a parabola of the form \(y = (x + p)^{2} + q\).

First use the information provided in the graph of the inverse:

\begin{align*} \text{Turning point: } & (3;1) \\ x-\text{intercept: } & (1;0) \end{align*}

To get the turning point and intercepts of the function, we invert the given coordinates. Now we can use those coordinates to find the equation of the function:

Now find the equation of the function:

\begin{align*} \text{Turning point: } & (1;3) \\ x-\text{intercept: } & (0;1) \\ y & = a(x-p)^{2} + q \\ y & = a(x-1)^{2} + 3 \\ \text{Substitute } (0;1) \quad 1 & = a(0-1)^{2} + 3\\ a & = -\text{2}\\ \therefore y & = -2(x-1)^{2} + 3 \end{align*}

Will \(f\) have a maximum or a minimum value?

Maximum value at \((1;3)\)

State the domain, range and axis of symmetry of \(f\).

Domain: \(\{x: x \in \mathbb{R} \}\) and range: \(\{y: y \leq 3,y \in \mathbb{R} \}\), Axis of symmetry: \(x =1\).

If \((q;3)\) lies on \(k\), determine the value(s) of \(q\).

\begin{align*} k(x) & = 2x^{2} + 1 \\ \text{Substitute } (q;3) \quad 3 & = 2(q)^{2} + 1 \\ 2 & = 2(q)^{2} \\ 1 & = q^{2} \\ \therefore q & = \pm 1 \end{align*}

This gives the points \((-1;3)\) and \((1;3)\).

Sketch the graph of \(k\), label the point(s) \((q;3)\) on the graph.

Find the equation of the inverse of \(k\) in the form \(y = \ldots\)

\begin{align*} k: \quad y & = 2x^{2} + 1 \\ \text{Inverse: } \quad x & = 2y^{2} + 1 \\ 2y^{2} & = x - 1 \\ y^{2} & = \frac{x-1}{2} \\ y & = \pm \sqrt{\frac{x-1}{2}} \qquad (x \ge 1) \end{align*}

Sketch \(k\) and \(y = \sqrt{\frac{x-1}{2}}\) on the same system of axes.

Determine the coordinates of the point on the graph of the inverse that is symmetrical to \((q;3)\) about the line \(y=x\).

\((3;1)\)

Determine the equation of \(f\).

\begin{align*} q & = -2 \\ y & = ax^{2} - 2 \\ \text{Substitute } (-2;-6) \quad -6 & = a(-2)^{2} - 2 \\ -6 + 2 & = 4a \\ -4 & = 4a \\ -1 & = a \\ \therefore f(x) & = -x^{2} - 2 \end{align*}

Determine and investigate the inverse.

\begin{align*} \text{Let } y & = -x^{2} - 2 \qquad (y \leq -2) \\ \text{Interchange } x \text{ and } y: \quad x & = -y^{2} - 2 \qquad (x \leq -2) \\ x + 2 & = -y^{2} \\ -x - 2 & = y^{2} \\ y & = \pm \sqrt{-x - 2} \qquad (x \leq -2) \end{align*}

Sketch the inverse and discuss the characteristics of the graph.

The inverse is a not a function. The turning point of the inverse is \((-2;0)\) and \(x\)-intercept is \((-2;0)\).

\[\text{Inverse}: \qquad \text{domain } \{x: x \le -2, x \in \mathbb{R} \} \quad \text{range } \{y: y \in \mathbb{R} \}\]

Determine the algebraic formula for the inverse of \(H\).

\begin{align*} \text{Let } y & = x^{2} - 9 \qquad (y \geq -9) \\ \text{Interchange } x \text{ and } y: \quad x & = y^{2} - 9 \qquad (x \geq -9) \\ x + 9 & = y^{2} \\ y & = \pm \sqrt{x + 9} \qquad (x \geq -9) \end{align*}

Draw graphs of \(H\) and its inverse on the same system of axes. Indicate intercepts and turning points.

Determine the intercepts:

\begin{align*} \text{Let } x = 0: \quad y & = (0)^{2} - 9 \\ & = - 9 \\ \text{Let } y = 0: \quad 0 & = x^{2} - 9 \\ x^2 & = 9 \\ \therefore x & = \pm 3 \end{align*}

The intercepts are \((0;-9)\) and \((-3;0),(3;0)\).

Is the inverse a function? Give reasons.

No. The inverse does not pass the vertical line test. It is a one-to-many relation.

Show algebraically and graphically the effect of restricting the domain of \(H\) to \(\{x: x \le 0 \}\).

If the domain of \(H\) is restricted to \(\{x: x \le 0 \}\), then the inverse is \(H^{-1}(x) = - \sqrt{x^{2} + 9} \quad (x \ge -9, y \le 0)\).

The graph of \(H^{-1}\) cuts a vertical line only once at any one time and therefore passes the vertical line test.