1.02u Functions: definition and vocabulary (domain, range, mapping)

15 Functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined as follows. \(\mathrm { f } ( x ) = \sqrt { x }\) for \(x > 0\) and \(\mathrm { g } ( x ) = x ^ { 3 } - x - 6\) for \(x > 2\). The function \(\mathrm { h } ( x )\) is defined as \(\mathrm { h } ( x ) = \mathrm { fg } ( x )\).

1. Find \(\mathrm { h } ( x )\) in terms of \(x\) and state its domain.
2. Find \(\mathrm { h } ( 3 )\). Fig. 15 shows \(\mathrm { h } ( x )\) and \(\mathrm { h } ^ { - 1 } ( x )\), together with the straight line \(y = x\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{cea67565-8074-4703-8e1a-09b98e380baf-17_780_796_895_242} \captionsetup{labelformat=empty} \caption{Fig. 15}
   \end{figure}
3. Determine the gradient of \(\mathrm { y } = \mathrm { h } ^ { - 1 } ( \mathrm { x } )\) at the point where \(y = 3\).

2 The function \(\mathrm { f } ( x ) = \sqrt { x }\) is defined on the domain \(x \geqslant 0\).
The function \(\mathrm { g } ( x ) = 25 - x ^ { 2 }\) is defined on the domain \(\mathbb { R }\).

1. Write down an expression for \(\mathrm { fg } ( x )\).
2. 1. Find the domain of \(\mathrm { fg } ( x )\).
   2. Find the range of \(\mathrm { fg } ( x )\).

2

1. The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = \sqrt { 1 + 2 x } \text { for } x \geqslant - \frac { 1 } { 2 }$$ Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of this inverse function.
2. Explain why \(\mathrm { g } ( x ) = 1 + x ^ { 2 }\), with domain all real numbers, has no inverse function.

3 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = 3 - x ^ { 2 } , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 2 } { x + 1 } , & \text { for real values of } x , x \neq - 1 \end{array}$$

1. Find the range of f.
2. The inverse of g is \(\mathrm { g } ^ { - 1 }\).
   1. Find \(\mathrm { g } ^ { - 1 } ( x )\).
   2. State the range of \(\mathrm { g } ^ { - 1 }\).
3. The composite function gf is denoted by h .
   1. Find \(\mathrm { h } ( x )\), simplifying your answer.
   2. State the greatest possible domain of h .

4 The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x ^ { 3 } , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 1 } { x - 3 } , & \text { for real values of } x , x \neq 3 \end{array}$$

1. State the range of f.
2. 1. Find fg(x).
   2. Solve the equation \(\operatorname { fg } ( x ) = 64\).
3. 1. The inverse of g is \(\mathrm { g } ^ { - 1 }\). Find \(\mathrm { g } ^ { - 1 } ( x )\).
   2. State the range of \(\mathrm { g } ^ { - 1 }\).

4 The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = 3 \cos \frac { 1 } { 2 } x , & \text { for } 0 \leqslant x \leqslant 2 \pi \\ \mathrm {~g} ( x ) = | x | , & \text { for all real values of } x \end{array}$$

1. Find the range of f .
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = 1\), giving your answer in an exact form.
3. 1. Write down an expression for \(\mathrm { gf } ( x )\).
   2. Sketch the graph of \(y = \operatorname { gf } ( x )\) for \(0 \leqslant x \leqslant 2 \pi\).
4. Describe a sequence of two geometrical transformations that maps the graph of \(y = \cos x\) onto the graph of \(y = 3 \cos \frac { 1 } { 2 } x\).

2 The curve with equation \(y = \frac { 63 } { 4 x - 1 }\) is sketched below for \(1 \leqslant x \leqslant 16\). \includegraphics[max width=\textwidth, alt={}, center]{7aa76d26-e3c4-4374-ae4f-8bb61e61b135-2_568_698_1308_669} The function f is defined by \(\mathrm { f } ( x ) = \frac { 63 } { 4 x - 1 }\) for \(1 \leqslant x \leqslant 16\).

1. Find the range of f .
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = 1\).
3. The function g is defined by \(\mathrm { g } ( x ) = x ^ { 2 }\) for \(- 4 \leqslant x \leqslant - 1\).
   1. Write down an expression for \(\mathrm { fg } ( x )\).
   2. Solve the equation \(\operatorname { fg } ( x ) = 1\).

5 The function f is defined by $$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 4 } { 3 } , \text { for real values of } x , \text { where } \boldsymbol { x } \leqslant \mathbf { 0 }$$

1. State the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Write down the domain of \(\mathrm { f } ^ { - 1 }\).
   2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. The function g is defined by $$\mathrm { g } ( x ) = \ln | 3 x - 1 | , \quad \text { for real values of } x , \text { where } x \neq \frac { 1 } { 3 }$$ The curve with equation \(y = \mathrm { g } ( x )\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-6_469_819_1254_612}
   1. The curve \(y = \mathrm { g } ( x )\) intersects the \(x\)-axis at the origin and at the point \(P\). Find the \(x\)-coordinate of \(P\).
   2. State whether the function \(g\) has an inverse. Give a reason for your answer.
   3. Show that \(\operatorname { gf } ( x ) = \ln \left| x ^ { 2 } - k \right|\), stating the value of the constant \(k\).
   4. Solve the equation \(\mathrm { gf } ( x ) = 0\).

2 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x - 2 & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 6 } { x + 3 } & \text { for real values of } x , \quad x \neq - 3 \end{array}$$ The composite function fg is denoted by h .

1. Find \(\mathrm { h } ( x )\).
2. 1. Find \(\mathrm { h } ^ { - 1 } ( x )\), where \(\mathrm { h } ^ { - 1 }\) is the inverse of h .
   2. Find the range of \(\mathrm { h } ^ { - 1 }\).

4 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x ^ { 2 } , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 1 } { 2 x - 3 } , & \text { for real values of } x , x \neq \frac { 3 } { 2 } \end{array}$$

1. State the range of f.
2. 1. The inverse of g is \(\mathrm { g } ^ { - 1 }\). Find \(\mathrm { g } ^ { - 1 } ( x )\).
   2. State the range of \(\mathrm { g } ^ { - 1 }\).
3. Solve the equation \(\operatorname { fg } ( x ) = 9\).

2 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } f ( x ) = \sqrt { 2 x + 5 } , & \text { for real values of } x , x \geqslant - 2.5 \\ g ( x ) = \frac { 1 } { 4 x + 1 } , & \text { for real values of } x , x \neq - 0.25 \end{array}$$

1. Find the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   2. State the domain of \(\mathrm { f } ^ { - 1 }\).
3. The composite function fg is denoted by h .
   1. Find an expression for \(\mathrm { h } ( x )\).
   2. Solve the equation \(\mathrm { h } ( x ) = 3\).

3

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when:
   1. \(\quad y = \ln ( 5 x - 2 )\);
   2. \(y = \sin 2 x\).
2. The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = \ln ( 5 x - 2 ) , & \text { for real values of } x \text { such that } x \geqslant \frac { 1 } { 2 } \\ \mathrm {~g} ( x ) = \sin 2 x , & \text { for real values of } x \text { in the interval } - \frac { \pi } { 4 } \leqslant x \leqslant \frac { \pi } { 4 } \end{array}$$
   1. Find the range of f .
   2. Find an expression for \(\operatorname { gf } ( x )\).
   3. Solve the equation \(\operatorname { gf } ( x ) = 0\).
   4. The inverse of g is \(\mathrm { g } ^ { - 1 }\). Find \(\mathrm { g } ^ { - 1 } ( x )\).

5 The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = \sqrt { 2 x - 5 } , & \text { for } x \geqslant 2.5 \\ \mathrm {~g} ( x ) = \frac { 10 } { x } , & \text { for real values of } x , \quad x \neq 0 \end{array}$$

1. State the range of f .
2. 1. Find \(\mathrm { fg } ( x )\).
   2. Solve the equation \(\operatorname { fg } ( x ) = 5\).
3. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = 7\).

5 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x ^ { 2 } - 6 x + 5 , & \text { for } x \geqslant 3 \\ \mathrm {~g} ( x ) = | x - 6 | , & \text { for all real values of } x \end{array}$$

1. Find the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\). Find \(\mathrm { f } ^ { - 1 } ( x )\). Give your answer in its simplest form.
3. 1. Find \(\mathrm { gf } ( x )\).
   2. Solve the equation \(\operatorname { gf } ( x ) = 6\).

5 The function f is defined by $$\mathrm { f } ( x ) = 16 x - \mathrm { e } ^ { 2 x } , \text { for all real } x$$ The graph of \(y = \mathrm { f } ( x )\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{bf427498-f1ee-4167-a6f2-ddaa2ff5ef81-12_789_1349_534_347}

1. Find the range of f.
2. The composite function fg is defined by $$\operatorname { fg } ( x ) = \frac { 16 } { x } - \mathrm { e } ^ { \frac { 2 } { x } } , \text { for real } x , x \neq 0$$ Find an expression for \(\operatorname { gg } ( x )\).

6. The function f is defined by $$\mathrm { f } ( x ) \equiv 4 - \ln 3 x , \quad x \in \mathbb { R } , \quad x > 0 .$$

1. Solve the equation \(\mathrm { f } ( x ) = 0\).
2. Sketch the curve \(y = \mathrm { f } ( x )\).
3. Find an expression for the inverse function, \(\mathrm { f } ^ { - 1 } ( x )\). The function \(g\) is defined by $$\mathrm { g } ( x ) \equiv \mathrm { e } ^ { 2 - x } , \quad x \in \mathbb { R }$$
4. Show that $$\operatorname { fg } ( x ) = x + a - \ln b$$ where \(a\) and \(b\) are integers to be found.

1. The function f is defined by

$$\mathrm { f } ( x ) \equiv 2 + \ln ( 3 x - 2 ) , \quad x \in \mathbb { R } , \quad x > \frac { 2 } { 3 } .$$

1. Find the exact value of \(\mathrm { ff } ( 1 )\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
   

6. The function f is defined by $$\mathrm { f } ( x ) \equiv 3 - x ^ { 2 } , \quad x \in \mathbb { R } , \quad x \geq 0 .$$

1. State the range of f.
2. Sketch the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) on the same diagram.
3. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain. The function g is defined by $$\mathrm { g } ( x ) \equiv \frac { 8 } { 3 - x } , \quad x \in \mathbb { R } , \quad x \neq 3 .$$
4. Evaluate \(\mathrm { fg } ( - 3 )\).
5. Solve the equation $$\mathrm { f } ^ { - 1 } ( x ) = \mathrm { g } ( x ) .$$

2. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \rightarrow 1 - a x , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow x ^ { 2 } + 2 a x + 2 , \quad x \in \mathbb { R } , \end{aligned}$$ where \(a\) is a constant.

1. Find the range of g in terms of \(a\). Given that \(\operatorname { gf } ( 3 ) = 7\),
2. find the two possible values of \(a\).

1. The functions \(f\) and \(g\) are defined by

$$\begin{aligned} & f : x \rightarrow 2 - x ^ { 2 } , \quad x \in \mathbb { R } , \\ & g : x \rightarrow \frac { 3 x } { 2 x - 1 } , \quad x \in \mathbb { R } , \quad x \neq \frac { 1 } { 2 } \end{aligned}$$

1. Evaluate fg(2).
2. Solve the equation \(\operatorname { gf } ( x ) = \frac { 1 } { 2 }\).
   

5. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } ( x ) \equiv 6 x - 1 , \quad x \in \mathbb { R } , \\ & \mathrm {~g} ( x ) \equiv \log _ { 2 } ( 3 x + 1 ) , \quad x \in \mathbb { R } , \quad x > - \frac { 1 } { 3 } \end{aligned}$$

1. Evaluate \(\operatorname { gf } ( 1 )\).
2. Find an expression for \(\mathrm { g } ^ { - 1 } ( x )\).
3. Find, in terms of natural logarithms, the solution of the equation $$\mathrm { fg } ^ { - 1 } ( x ) = 2$$

7. The function f is defined by $$\mathrm { f } ( x ) \equiv x ^ { 2 } - 2 a x , \quad x \in \mathbb { R } ,$$ where \(a\) is a positive constant.

1. Showing the coordinates of any points where each graph meets the axes, sketch on separate diagrams the graphs of
   1. \(\quad y = | \mathrm { f } ( x ) |\),
   2. \(y = \mathrm { f } ( | x | )\). The function g is defined by $$\mathrm { g } ( x ) \equiv 3 a x , \quad x \in \mathbb { R } .$$
2. Find fg(a) in terms of \(a\).
3. Solve the equation $$\operatorname { gf } ( x ) = 9 a ^ { 3 }$$

2. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } ( x ) \equiv x ^ { 2 } - 3 x + 7 , \quad x \in \mathbb { R } \\ & \mathrm {~g} ( x ) \equiv 2 x - 1 , \quad x \in \mathbb { R } \end{aligned}$$

1. Find the range of f .
2. Evaluate \(\operatorname { gf } ( - 1 )\).
3. Solve the equation $$\mathrm { fg } ( x ) = 17$$
   1. \(f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 13 x ^ { 2 } + 26 x - 17 } { x ^ { 2 } - 3 x + 3 } , x \in \mathbb { R }\).
   2. Find the values of the constants \(A\), \(B\), \(C\) and \(D\) such that
   $$f ( x ) = x ^ { 2 } + A x + B + \frac { C x + D } { x ^ { 2 } - 3 x + 3 }$$ The point \(P\) on the curve \(y = \mathrm { f } ( x )\) has \(x\)-coordinate 1.
4. Show that the normal to the curve \(y = \mathrm { f } ( x )\) at \(P\) has the equation $$x + 5 y + 9 = 0$$
   1. (a) Given that
   $$x = \sec \frac { y } { 2 } , \quad 0 \leq y < \pi ,$$ show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x \sqrt { x ^ { 2 } - 1 } } .$$
5. Find an equation for the tangent to the curve \(y = \sqrt { 3 + 2 \cos x }\) at the point where \(x = \frac { \pi } { 3 }\).

7. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$

1. State the range of f .
2. Evaluate fg(-2).
3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
5. Use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures. \end{document}

4 \includegraphics[max width=\textwidth, alt={}, center]{28beb431-45d5-4300-88fe-00d05d78790b-05_787_892_267_568} The diagram shows the graph of \(y = \mathrm { f } ( x )\), where $$f ( x ) = \begin{cases} 4 - 4 x , & x \leqslant a , \\ \ln ( b x - 8 ) - 2 , & x \geqslant a . \end{cases}$$ The range of f is \(\mathrm { f } ( x ) \geqslant - 2\).

1. Show that \(a = \frac { 3 } { 2 }\).
2. Find the value of \(b\).
3. Find the exact value of \(\mathrm { ff } ( - 1 )\).
4. Explain why the function f does not have an inverse.