1.02v Inverse and composite functions: graphs and conditions for existence

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

$$\begin{array} { l l } \mathrm { f } : x \mapsto \mathrm { e } ^ { - x } + 2 , & x \in \mathbb { R } \\ \mathrm {~g} : x \mapsto 2 \ln x , & x > 0 \end{array}$$

1. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
2. Find the exact value of \(x\) for which \(\mathrm { f } ( 2 x + 3 ) = 6\)
3. Find \(\mathrm { f } ^ { - 1 }\), stating its domain.
4. 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.
   

8. The functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } : x \rightarrow 2 x + \ln 2 , & x \in \mathbb { R } , \\ \mathrm {~g} : x \rightarrow \mathrm { e } ^ { 2 x } , & x \in \mathbb { R } . \end{array}$$

1. Prove that the composite function gf is $$\operatorname { gf } : x \rightarrow 4 \mathrm { e } ^ { 4 x } , \quad x \in \mathbb { R }$$
2. In the space provided on page 19, sketch the curve with equation \(y = \operatorname { gf } ( x )\), and show the coordinates of the point where the curve cuts the \(y\)-axis.
3. Write down the range of gf.
4. Find the value of \(x\) for which \(\frac { \mathrm { d } } { \mathrm { d } x } [ \operatorname { gf } ( x ) ] = 3\), giving your answer to 3 significant figures.

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

$$\mathrm { f } : x \mapsto \ln ( 4 - 2 x ) , \quad x < 2 \quad \text { and } \quad x \in \mathbb { R } .$$

1. Show that the inverse function of f is defined by $$\mathrm { f } ^ { - 1 } : x \mapsto 2 - \frac { 1 } { 2 } \mathrm { e } ^ { x }$$ and write down the domain of \(\mathrm { f } ^ { - 1 }\).
2. Write down the range of \(\mathrm { f } ^ { - 1 }\).
3. In the space provided on page 16, sketch the graph of \(y = f ^ { - 1 } ( x )\). State the coordinates of the points of intersection with the \(x\) and \(y\) axes. The graph of \(y = x + 2\) crosses the graph of \(y = f ^ { - 1 } ( x )\) at \(x = k\). The iterative formula $$x _ { n + 1 } = - \frac { 1 } { 2 } e ^ { x _ { n } } , x _ { 0 } = - 0.3$$ is used to find an approximate value for \(k\).
4. Calculate the values of \(x _ { 1 }\) and \(x _ { 2 }\), giving your answers to 4 decimal places.
5. Find the value of \(k\) to 3 decimal places.

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

$$\begin{aligned} & \mathrm { f } : x \mapsto 1 - 2 x ^ { 3 } , x \in \mathbb { R } \\ & \mathrm {~g} : x \mapsto \frac { 3 } { x } - 4 , x > 0 , x \in \mathbb { R } \end{aligned}$$

1. Find the inverse function \(\mathrm { f } ^ { - 1 }\).
2. Show that the composite function gf is $$\text { gf } : x \mapsto \frac { 8 x ^ { 3 } - 1 } { 1 - 2 x ^ { 3 } }$$
3. Solve \(\operatorname { gf } ( x ) = 0\).
4. Use calculus to find the coordinates of the stationary point on the graph of \(y = \operatorname { gf } ( x )\).

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 \(\mathrm { g } ^ { - 1 }\).

1. The function f is defined by

$$\mathrm { f } : x \mapsto \frac { 3 ( x + 1 ) } { 2 x ^ { 2 } + 7 x - 4 } - \frac { 1 } { x + 4 } , \quad x \in \mathbb { R } , x > \frac { 1 } { 2 }$$

1. Show that \(\mathrm { f } ( x ) = \frac { 1 } { 2 x - 1 }\)
2. Find \(\mathrm { f } ^ { - 1 } ( x )\)
3. Find the domain of \(\mathrm { f } ^ { - 1 }\) $$\mathrm { g } ( x ) = \ln ( x + 1 )$$
4. Find the solution of \(\mathrm { fg } ( x ) = \frac { 1 } { 7 }\), giving your answer in terms of e .

3. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\).
The curve passes through the points \(Q ( 0,2 )\) and \(P ( - 3,0 )\) as shown.

1. Find the value of ff(-3). On separate diagrams, sketch the curve with equation
2. \(y = \mathrm { f } ^ { - 1 } ( x )\),
3. \(y = \mathrm { f } ( | x | ) - 2\),
4. \(y = 2 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\). Indicate clearly on each sketch the coordinates of the points at which the curve crosses or meets the axes.

8. The function \(f\) is defined by $$\mathrm { f } : x \rightarrow 3 - 2 \mathrm { e } ^ { - x } , \quad x \in \mathbb { R }$$

1. Find the inverse function, \(\mathrm { f } ^ { - 1 } ( x )\) and give its domain.
2. Solve the equation \(\mathrm { f } ^ { - 1 } ( x ) = \ln x\). The equation \(\mathrm { f } ( t ) = k \mathrm { e } ^ { t }\), where \(k\) is a positive constant, has exactly one real solution.
3. Find the value of \(k\).

3. The function \(f\) is defined by $$f : x \rightarrow \frac { 5 x + 1 } { x ^ { 2 } + x - 2 } - \frac { 3 } { x + 2 } , x > 1$$

1. Show that \(\mathrm { f } ( x ) = \frac { 2 } { x - 1 } , x > 1\).
2. Find \(\mathrm { f } ^ { - 1 } ( x )\). The function \(g\) is defined by $$\mathrm { g } : x \rightarrow x ^ { 2 } + 5 , \quad x \in \mathbb { R }$$
3. Solve \(\operatorname { fg } ( x ) = \frac { 1 } { 4 }\).

7. For the constant \(k\), where \(k > 1\), the functions f and g are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto \ln ( x + k ) , \quad x > - k , \\ & \mathrm {~g} : x \mapsto | 2 x - k | , \quad x \in \mathbb { R } . \end{aligned}$$

1. On separate axes, sketch the graph of f and the graph of g . On each sketch state, in terms of \(k\), the coordinates of points where the graph meets the coordinate axes.
2. Write down the range of f.
3. Find \(\mathrm { fg } \left( \frac { k } { 4 } \right)\) in terms of \(k\), giving your answer in its simplest form. The curve \(C\) has equation \(y = \mathrm { f } ( x )\). The tangent to \(C\) at the point with \(x\)-coordinate 3 is parallel to the line with equation \(9 y = 2 x + 1\).
4. Find the value of \(k\).

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 $$\operatorname { gg } ( x ) + [ \mathrm { g } ( x ) ] ^ { 2 } = 0$$

1. $$g ( x ) = \frac { 6 x + 12 } { x ^ { 2 } + 3 x + 2 } - 2 , \quad x \geqslant 0$$

1. Show that \(\mathrm { g } ( x ) = \frac { 4 - 2 x } { x + 1 } , x \geqslant 0\)
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0f6fd881-4d4b-4f80-96cc-6da41cc33c60-02_494_922_628_511} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { g } ( x ) , x \geqslant 0\) The curve meets the \(y\)-axis at \(( 0,4 )\) and crosses the \(x\)-axis at \(( 2,0 )\). On separate diagrams sketch the graph with equation
   1. \(y = 2 \mathrm {~g} ( 2 x )\),
   2. \(y = \mathrm { g } ^ { - 1 } ( x )\). Show on each sketch the coordinates of each point at which the graph meets or crosses the axes.

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 \(\operatorname { gf } ( x ) = 16\)

6. The function f is defined by $$\mathrm { f } : x \rightarrow \mathrm { e } ^ { 2 x } + k ^ { 2 } , \quad x \in \mathbb { R } , \quad k \text { is a positive constant. }$$

1. State the range of f .
2. Find \(\mathrm { f } ^ { - 1 }\) and state its domain. The function g is defined by $$g : x \rightarrow \ln ( 2 x ) , \quad x > 0$$
3. Solve the equation $$\mathrm { g } ( x ) + \mathrm { g } \left( x ^ { 2 } \right) + \mathrm { g } \left( x ^ { 3 } \right) = 6$$ giving your answer in its simplest form.
4. Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
5. Find, in terms of the constant \(k\), the solution of the equation $$\mathrm { fg } ( x ) = 2 k ^ { 2 }$$

5. $$\mathrm { g } ( x ) = \frac { x } { x + 3 } + \frac { 3 ( 2 x + 1 ) } { x ^ { 2 } + x - 6 } , \quad x > 3$$

1. Show that \(\mathrm { g } ( x ) = \frac { x + 1 } { x - 2 } , \quad x > 3\)
2. Find the range of g.
3. Find the exact value of \(a\) for which \(\mathrm { g } ( a ) = \mathrm { g } ^ { - 1 } ( a )\).

7. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation $$\mathrm { g } ( x ) = x ^ { 2 } ( 1 - x ) \mathrm { e } ^ { - 2 x } , \quad x \geqslant 0$$

1. Show that \(\mathrm { g } ^ { \prime } ( x ) = \mathrm { f } ( x ) \mathrm { e } ^ { - 2 x }\), where \(\mathrm { f } ( x )\) is a cubic function to be found.
2. Hence find the range of g .
3. State a reason why the function \(\mathrm { g } ^ { - 1 } ( x )\) does not exist.