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

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.

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

$$\begin{aligned} & \mathrm { f } : x \rightarrow 7 x - 1 , \quad x \in \mathbb { R } \\ & \mathrm {~g} : x \rightarrow \frac { 4 } { x - 2 } , \quad x \neq 2 , x \in \mathbb { R } \end{aligned}$$

1. Solve the equation \(\operatorname { fg } ( x ) = x\)
2. Hence, or otherwise, find the largest value of \(a\) such that \(\mathrm { g } ( a ) = \mathrm { f } ^ { - 1 } ( a )\)

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the graph of \(y = \mathrm { g } ( x )\), where $$g ( x ) = 3 + \sqrt { x + 2 } , \quad x \geqslant - 2$$

1. State the range of g .
2. Find \(\mathrm { g } ^ { - 1 } ( x )\) and state its domain.
3. Find the exact value of \(x\) for which $$\mathrm { g } ( x ) = x$$
4. Hence state the value of \(a\) for which $$\mathrm { g } ( a ) = \mathrm { g } ^ { - 1 } ( a )$$

1. The function f is defined by

$$\mathrm { f } ( x ) = \frac { 6 } { 2 x + 5 } + \frac { 2 } { 2 x - 5 } + \frac { 60 } { 4 x ^ { 2 } - 25 } , \quad x > 4$$

1. Show that \(\mathrm { f } ( x ) = \frac { A } { B x + C }\) where \(A , B\) and \(C\) are constants to be found.
2. Find \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain.

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

1. Find \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\). The function g is defined by $$\mathrm { g } : x \mapsto 3 x ^ { 2 } + 2 , \quad x \in \mathbb { R }$$
2. Find \(\mathrm { gf } ^ { - 1 } ( x )\).
3. State the range of \(\mathrm { gf } ^ { - 1 } ( x )\).

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve \(C\) with parametric equations $$x = 2 \cos 2 t \quad y = 4 \sin t \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$ The region \(R\), shown shaded in Figure 3, is bounded by the curve, the \(x\)-axis and the \(y\)-axis.

1. 1. Show, making your working clear, that the area of \(R = \int _ { 0 } ^ { \frac { \pi } { 4 } } 32 \sin ^ { 2 } t \cos t d t\)
   2. Hence find, by algebraic integration, the exact value of the area of \(R\).
2. Show that all points on \(C\) satisfy \(y = \sqrt { a x + b }\), where \(a\) and \(b\) are constants to be found. The curve \(C\) has equation \(y = \mathrm { f } ( x )\) where f is the function $$f ( x ) = \sqrt { a x + b } \quad - 2 \leqslant x \leqslant 2$$ and \(a\) and \(b\) are the constants found in part (b).
3. State the range of f.

2 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for all real numbers \(x\) by $$\mathrm { f } ( x ) = x ^ { 2 } , \quad \mathrm {~g} ( x ) = x - 2$$

1. Find the composite functions \(\mathrm { fg } ( x )\) and \(\mathrm { gf } ( x )\).
2. Sketch the curves \(y = \mathrm { f } ( x ) , y = \mathrm { fg } ( x )\) and \(y = \mathrm { gf } ( x )\), indicating clearly which is which.

3 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for the domain \(x > 0\) as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function \(\mathrm { fg } ( x )\) in terms of \(\ln x\).
State the transformation which maps the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { fg } ( x )\).

9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
4. Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(\mathrm { g } ( x )\). Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

6. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \rightarrow 3 x - 4 , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \frac { 2 } { x + 3 } , \quad x \in \mathbb { R } , \quad x \neq - 3 \end{aligned}$$

1. Evaluate fg(1).
2. Solve the equation \(\operatorname { gf } ( x ) = 6\).
3. Find an expression for \(\mathrm { g } ^ { - 1 } ( x )\).

9. \(\quad f ( x ) = 3 - e ^ { 2 x } , \quad x \in \mathbb { R }\).

1. State the range of f .
2. Find the exact value of \(\mathrm { ff } ( 0 )\).
3. Define the inverse function \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain. Given that \(\alpha\) is the solution of the equation \(\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )\),
4. explain why \(\alpha\) satisfies the equation $$x = \mathrm { f } ^ { - 1 } ( x )$$
5. use the iterative formula $$x _ { n + 1 } = \mathrm { f } ^ { - 1 } \left( x _ { n } \right)$$ with \(x _ { 0 } = 0.5\) to find \(\alpha\) correct to 3 significant figures.

7. The function \(f\) is defined by $$\mathrm { f } : x \rightarrow 3 \mathrm { e } ^ { x - 1 } , \quad x \in \mathbb { R }$$

1. State the range of f .
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain. The function \(g\) is defined by $$g : x \rightarrow 5 x - 2 , \quad x \in \mathbb { R }$$ Find, in terms of e,
3. the value of \(\mathrm { gf } ( \ln 2 )\),
4. the solution of the equation $$\mathrm { f } ^ { - 1 } \mathrm {~g} ( x ) = 4$$

2. 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 \(\operatorname { gf } ( x ) = \frac { 1 } { 2 }\).

4. 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 the graph meets the axes, sketch the graph of \(y = | \mathrm { f } ( x ) |\). The function \(g\) is defined by $$\mathrm { g } ( x ) \equiv 3 a x , \quad x \in \mathbb { R } .$$
2. Find \(\mathrm { fg } ( \mathrm { a } )\) in terms of \(a\).
3. Solve the equation $$\operatorname { gf } ( x ) = 9 a ^ { 3 }$$

4 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-2_529_737_900_701} The function f is defined by \(\mathrm { f } ( x ) = 2 - \sqrt { x }\) for \(x \geqslant 0\). The graph of \(y = \mathrm { f } ( x )\) is shown above.

1. State the range of f.
2. Find the value of \(\mathrm { ff } ( 4 )\).
3. Given that the equation \(| \mathrm { f } ( x ) | = k\) has two distinct roots, determine the possible values of the constant \(k\).

9 Functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = 2 \sin x & \text { for } - \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi \\ \mathrm {~g} ( x ) = 4 - 2 x ^ { 2 } & \text { for } x \in \mathbb { R } . \end{array}$$

1. State the range of f and the range of g .
2. Show that \(\operatorname { gf } ( 0.5 ) = 2.16\), correct to 3 significant figures, and explain why \(\mathrm { fg } ( 0.5 )\) is not defined.
3. Find the set of values of \(x\) for which \(\mathrm { f } ^ { - 1 } \mathrm {~g} ( x )\) is not defined.

1 The function f is defined for all real values of \(x\) by $$f ( x ) = 10 - ( x + 3 ) ^ { 2 } .$$

1. State the range of f .
2. Find the value of \(\mathrm { ff } ( - 1 )\).

6 \includegraphics[max width=\textwidth, alt={}, center]{ebfdf170-99c6-4785-b9d7-201c3425b4c9-3_563_583_267_781} The diagram shows the graph of \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = 2 - x ^ { 2 } , \quad x \leqslant 0 .$$

1. Evaluate \(\mathrm { ff } ( - 3 )\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. Sketch the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\). Indicate the coordinates of the points where the graph meets the axes.

3 The function \(f\) is defined for all non-negative values of \(x\) by $$f ( x ) = 3 + \sqrt { x }$$

1. Evaluate ff(169).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( \mathrm { x } )\) in terms of x .
3. On a single diagram sketch the graphs of \(y = f ( x )\) and \(y = f ^ { - 1 } ( x )\), indicating how the two graphs are related.

9 \includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-4_534_935_264_605} The function f is defined for the domain \(x \geqslant 0\) by $$f ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$ The diagram shows the curve with equation \(y = \mathrm { f } ( x )\).

1. Find the range of f .
2. The function g is defined for the domain \(x \geqslant k\) by $$\mathrm { g } ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$ Given that g is a one-one function, state the least possible value of \(k\).
3. Show that there is no point on the curve \(y = \mathrm { g } ( x )\) at which the gradient is - 1 .

6 The function f is defined by $$\mathrm { f } : x \mapsto 1 + \sqrt { } x \quad \text { for } x \geqslant 0$$

1. State the domain and range of the inverse function \(\mathrm { f } ^ { - 1 }\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. By considering the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), show that the solution to the equation $$\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )$$ is \(x = \frac { 1 } { 2 } ( 3 + \sqrt { } 5 )\).

6 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 + 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x - 1 } { 2 } \right)\) and state the domain of this function. Fig. 6 shows a sketch of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-4_506_561_705_751} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .

2 Given that \(\mathrm { f } ( x ) = 1 - x\) and \(\mathrm { g } ( x ) = | x |\), write down the composite function \(\mathrm { gf } ( x )\).
On separate diagrams, sketch the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { gf } ( x )\).