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

1. The function f is defined by

$$\mathrm { f } ( x ) = \frac { 5 x } { x ^ { 2 } + 7 x + 12 } + \frac { 5 x } { x + 4 } \quad x > 0$$

1. Show that \(\mathrm { f } ( x ) = \frac { 5 x } { x + 3 }\)
2. Find \(\mathrm { f } ^ { - 1 }\)
3. 1. Find, in simplest form, \(\mathrm { f } ^ { \prime } ( x )\).
   2. Hence, state whether f is an increasing or a decreasing function, giving a reason for your answer.
      (3)

2. The functions f and g are defined by $$\begin{array} { l l } f ( x ) = 5 - \frac { 4 } { 3 x + 2 } & x \geqslant 0 \\ g ( x ) = \left| 4 \sin \left( \frac { x } { 3 } + \frac { \pi } { 6 } \right) \right| & x \in \mathbb { R } \end{array}$$

1. Find the range of f
2. 1. Find \(\mathrm { f } ^ { - 1 } ( x )\)
   2. Write down the domain of \(\mathrm { f } ^ { - 1 }\)
3. Find \(\mathrm { fg } ( - \pi )\)

1. The function f is defined by

$$\mathrm { f } ( x ) = \frac { x + 3 } { x - 4 } \quad x \in \mathbb { R } , x \neq 4$$

1. Find ff(6)
2. Find \(f ^ { - 1 }\) The function \(g\) is defined by $$g ( x ) = x ^ { 2 } + 5 \quad x \in \mathbb { R } , x > 0$$
3. Find the exact value of \(a\) for which $$\operatorname { gf } ( a ) = 7$$

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = e ^ { - x ^ { 2 } } \left( 2 x ^ { 2 } - 3 \right) ^ { 2 }$$

1. Find the range of f
2. Show that $$\mathrm { f } ^ { \prime } ( x ) = 2 x \left( 2 x ^ { 2 } - 3 \right) \mathrm { e } ^ { - x ^ { 2 } } \left( A - B x ^ { 2 } \right)$$ where \(A\) and \(B\) are constants to be found. Given that the line \(y = k\), where \(k\) is a constant, \(k > 0\), intersects the curve at exactly two distinct points,
3. find the exact range of values of \(k\)

1. The functions f and g are defined by

$$\mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 \quad x \in \mathbb { R }$$ $$\mathrm { g } : x \mapsto \ln x \quad x > 0$$

1. State the range of f .
2. Find \(\mathrm { fg } ( x )\), giving \(y\) our answer in its simplest form.
3. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
4. Find \(\mathrm { f } ^ { - 1 }\) stating its domain.
5. On the same axes sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.

4. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the graph \(y = \mathrm { f } ( x )\), where $$f ( x ) = 2 | 3 - x | + 5 , \quad x \geqslant 0$$ Figure 2 shows a sketch of part of the graph \(y = \mathrm { g } ( x )\), where $$\operatorname { g } ( x ) = \frac { x + 9 } { 2 x + 3 } , \quad x \geqslant 0$$

1. Find the value of \(\mathrm { fg } ( 1 )\)
2. State the range of g
3. Find \(\mathrm { g } ^ { - 1 } ( x )\) and state its domain. Given that the equation \(\mathrm { f } ( x ) = k\), where \(k\) is a constant, has exactly two roots,
4. state the range of possible values of \(k\).

1. Given that

$$\begin{array} { l l } \mathrm { f } ( x ) = \frac { 4 } { 3 x + 5 } , & x > 0 \\ \mathrm {~g} ( x ) = \frac { 1 } { x } , & x > 0 \end{array}$$

1. state the range of f,
2. find \(\mathrm { f } ^ { - 1 } ( x )\),
3. find \(\mathrm { fg } ( x )\).
4. Show that the equation \(\mathrm { fg } ( x ) = \mathrm { gf } ( x )\) has no real solutions.

1. It is given that

$$\begin{gathered} \mathrm { f } ( x ) = \mathrm { e } ^ { - 2 x } \quad x \in \mathbb { R } \\ \mathrm {~g} ( x ) = \frac { x } { x - 3 } \quad x > 3 \end{gathered}$$

1. Sketch the graph of \(y = \mathrm { f } ( x )\), showing the coordinates of any points where the graph crosses the axes.
2. Find the range of g
3. Find \(\mathrm { g } ^ { - 1 } ( x )\), stating the domain of \(\mathrm { g } ^ { - 1 }\)
4. Using algebra, find the exact value of \(x\) for which \(\operatorname { fg } ( x ) = 3\)

3. The function f is defined by $$f : x \mapsto 2 x ^ { 2 } + 3 k x + k ^ { 2 } \quad x \in \mathbb { R } , - 4 k \leqslant x \leqslant 0$$ where \(k\) is a positive constant.

1. Find, in terms of \(k\), the range of f . The function g is defined by $$\mathrm { g } : x \mapsto 2 k - 3 x \quad x \in \mathbb { R }$$ Given that \(\operatorname { gf } ( - 2 ) = - 12\)
2. find the possible values of \(k\).

3. The function g is defined by $$g ( x ) = \frac { 6 x } { 2 x + 3 } \quad x > 0$$

1. Find the range of g .
2. Find \(\mathrm { g } ^ { - 1 } ( x )\) and state its domain.
3. Find the function \(\operatorname { gg } ( x )\), writing your answer as a single fraction in its simplest form.

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

$$\begin{array} { l l } \mathrm { f } : x \rightarrow \mathrm { e } ^ { 2 x } - 5 , & x \in \mathbb { R } \\ \mathrm {~g} : x \rightarrow \ln ( 3 x - 1 ) , & x \in \mathbb { R } , x > \frac { 1 } { 3 } \end{array}$$

1. Find \(\mathrm { f } ^ { - 1 }\) and state its domain.
2. Find \(\mathrm { fg } ( 3 )\), giving your answer in its simplest form.
   (ii) (a) Sketch the graph with equation $$y = | 4 x - a |$$ where \(a\) is a positive constant. State the coordinates of each point where the graph cuts or meets the coordinate axes. Given that $$| 4 x - a | = 9 a$$ where \(a\) is a positive constant,
3. find the possible values of $$| x - 6 a | + 3 | x |$$ giving your answers, in terms of \(a\), in their simplest form.

14. Given that $$y = \frac { \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$

1. show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A x ^ { 2 } + 12 } { x ^ { 4 } \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } \quad x > 2$$ where \(A\) is a constant to be found. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-48_593_1134_865_395} \captionsetup{labelformat=empty} \caption{Figure 4}
   \end{figure} Figure 4 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \frac { 24 \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$
2. Use your answer to part (a) to find the range of f.
3. State a reason why f-1 does not exist.

5. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto 3 x + \ln x , \quad x > 0 , \quad x \in \mathbb { R } \\ & \mathrm {~g} : x \mapsto \mathrm { e } ^ { x ^ { 2 } } , \quad x \in \mathbb { R } \end{aligned}$$

1. Write down the range of g.
2. Show that the composite function fg is defined by $$\mathrm { fg } : x \mapsto x ^ { 2 } + 3 \mathrm { e } ^ { x ^ { 2 } } , \quad x \in \mathbb { R } .$$
3. Write down the range of fg.
4. Solve the equation \(\frac { \mathrm { d } } { \mathrm { d } x } [ \mathrm { fg } ( x ) ] = x \left( x \mathrm { e } ^ { x ^ { 2 } } + 2 \right)\).

9. (i) Find the exact solutions to the equations

1. \(\ln ( 3 x - 7 ) = 5\)
2. \(3 ^ { x } \mathrm { e } ^ { 7 x + 2 } = 15\) (ii) The functions f and g are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } + 3 , & x \in \mathbb { R } \\ \mathrm {~g} ( x ) = \ln ( x - 1 ) , & x \in \mathbb { R } , x > 1 \end{array}$$
   1. Find \(\mathrm { f } ^ { - 1 }\) and state its domain.
   2. Find fg and state its range.

1. The function \(f\) is defined by

$$\mathrm { f } : x \mapsto \frac { 3 - 2 x } { x - 5 } , \quad x \in \mathbb { R } , x \neq 5$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   (3) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{3ff6824f-9fbf-4b5b-8bab-91332c549b36-10_901_1091_593_429} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} The function g has domain \(- 1 \leqslant x \leqslant 8\), and is linear from \(( - 1 , - 9 )\) to \(( 2,0 )\) and from \(( 2,0 )\) to \(( 8,4 )\). Figure 2 shows a sketch of the graph of \(y = \mathrm { g } ( x )\).
2. Write down the range of g.
3. Find \(\operatorname { gg } ( 2 )\).
4. Find \(\mathrm { fg } ( 8 )\).
5. On separate diagrams, sketch the graph with equation
   1. \(y = | \mathrm { g } ( x ) |\),
   2. \(y = \mathrm { g } ^ { - 1 } ( x )\). Show on each sketch the coordinates of each point at which the graph meets or cuts the axes.
6. State the domain of the inverse function \(\mathrm { g } ^ { - 1 }\).

5. The functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } : x \mapsto \ln ( 2 x - 1 ) , & x \in \mathbb { R } , x > \frac { 1 } { 2 } \\ \mathrm {~g} : x \mapsto \frac { 2 } { x - 3 } , & x \in \mathbb { R } , x \neq 3 \end{array}$$

1. Find the exact value of fg(4).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\), stating its domain.
3. Sketch the graph of \(y = | \mathrm { g } ( x ) |\). Indicate clearly the equation of the vertical asymptote and the coordinates of the point at which the graph crosses the \(y\)-axis.
4. Find the exact values of \(x\) for which \(\left| \frac { 2 } { x - 3 } \right| = 3\).

4. The function \(f\) is defined by $$f : x \mapsto \frac { 2 ( x - 1 ) } { x ^ { 2 } - 2 x - 3 } - \frac { 1 } { x - 3 } , \quad x > 3$$

1. Show that \(\mathrm { f } ( x ) = \frac { 1 } { x + 1 } , \quad x > 3\).
2. Find the range of f.
3. Find \(\mathrm { f } ^ { - 1 } ( x )\). State the domain of this inverse function. The function \(g\) is defined by $$\mathrm { g } : x \mapsto 2 x ^ { 2 } - 3 , \quad x \in \mathbb { R }$$
4. Solve \(\mathrm { fg } ( x ) = \frac { 1 } { 8 }\).

5. Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\).
The curve meets the coordinate axes at the points \(A ( 0,1 - k )\) and \(B \left( \frac { 1 } { 2 } \ln k , 0 \right)\), where \(k\) is a constant and \(k > 1\), as shown in Figure 2. On separate diagrams, sketch the curve with equation

1. \(y = | f ( x ) |\),
2. \(y = \mathrm { f } ^ { - 1 } ( x )\). Show on each sketch the coordinates, in terms of \(k\), of each point at which the curve meets or cuts the axes. Given that \(\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } - k\),
3. state the range of f ,
4. find \(\mathrm { f } ^ { - 1 } ( x )\),
5. write down the domain of \(\mathrm { f } ^ { - 1 }\).

4. The function \(f\) is defined by $$f : x \mapsto | 2 x - 5 | , \quad x \in \mathbb { R }$$

1. Sketch the graph with equation \(y = \mathrm { f } ( x )\), showing the coordinates of the points where the graph cuts or meets the axes.
2. Solve \(\mathrm { f } ( x ) = 15 + x\). The function \(g\) is defined by $$g : x \mapsto x ^ { 2 } - 4 x + 1 , \quad x \in \mathbb { R } , \quad 0 \leqslant x \leqslant 5$$
3. Find fg(2).
4. Find the range of g.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with the equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\). The curve has a turning point at \(A ( 3 , - 4 )\) and also passes through the point \(( 0,5 )\).

1. Write down the coordinates of the point to which \(A\) is transformed on the curve with equation
   1. \(y = | \mathrm { f } ( x ) |\),
   2. \(y = 2 f \left( \frac { 1 } { 2 } x \right)\).
2. Sketch the curve with equation $$y = \mathrm { f } ( | x | )$$ On your sketch show the coordinates of all turning points and the coordinates of the point at which the curve cuts the \(y\)-axis. The curve with equation \(y = \mathrm { f } ( x )\) is a translation of the curve with equation \(y = x ^ { 2 }\).
3. Find \(\mathrm { f } ( x )\).
4. Explain why the function f does not have an inverse.

4. The function \(f\) is defined by $$\mathrm { f } : x \mapsto 4 - \ln ( x + 2 ) , \quad x \in \mathbb { R } , x \geqslant - 1$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
2. Find the domain of \(\mathrm { f } ^ { - 1 }\). The function \(g\) is defined by $$\mathrm { g } : x \mapsto \mathrm { e } ^ { x ^ { 2 } } - 2 , \quad x \in \mathbb { R }$$
3. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
4. Find the range of fg.

6. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto \mathrm { e } ^ { x } + 2 , \quad x \in \mathbb { R } \\ & \mathrm {~g} : x \mapsto \ln x , \quad x > 0 \end{aligned}$$

1. State the range of f.
2. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
3. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
4. Find \(\mathrm { f } ^ { - 1 }\), the inverse function of f , stating its domain.
5. On the same axes sketch the curves with equation \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), giving the coordinates of all the points where the curves cross the axes.

1. The functions f and g are defined by

$$\begin{array} { l l } \mathrm { f } : x \mapsto 2 | x | + 3 , & x \in \mathbb { R } , \\ \mathrm {~g} : x \mapsto 3 - 4 x , & x \in \mathbb { R } \end{array}$$

1. State the range of f.
2. Find \(\mathrm { fg } ( 1 )\).
3. Find \(\mathrm { g } ^ { - 1 }\), the inverse function of g .
4. Solve the equation $$\operatorname { gg } ( x ) + [ \mathrm { g } ( x ) ] ^ { 2 } = 0$$

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the graph of \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \left\{ \begin{array} { r r } 5 - 2 x , & x \leqslant 4 \\ \mathrm { e } ^ { 2 x - 8 } - 4 , & x > 4 \end{array} \right.$$

1. State the range of \(\mathrm { f } ( x )\).
2. Determine the exact value of ff(0).
3. Solve \(\mathrm { f } ( x ) = 21\) Give each answer as an exact answer.
4. Explain why the function f does not have an inverse.

7. The function \(f\) has domain \(- 2 \leqslant x \leqslant 6\) and is linear from \(( - 2,10 )\) to \(( 2,0 )\) and from \(( 2,0 )\) to (6, 4). A sketch of the graph of \(y = \mathrm { f } ( x )\) is shown in Figure 1. \begin{figure}[h] \end{figure}

1. Write down the range of f .
2. Find \(\mathrm { ff } ( 0 )\). The function \(g\) is defined by $$\mathrm { g } : x \rightarrow \frac { 4 + 3 x } { 5 - x } , \quad x \in \mathbb { R } , \quad x \neq 5$$
3. Find \(\mathrm { g } ^ { - 1 } ( x )\)
4. Solve the equation \(\operatorname { gf } ( x ) = 16\)