Solve equation involving composites

8 Functions f and g are defined as follows: $$\begin{aligned} & \mathrm { f } : x \mapsto x ^ { 2 } - 1 \text { for } x < 0 \\ & \mathrm {~g} : x \mapsto \frac { 1 } { 2 x + 1 } \text { for } x < - \frac { 1 } { 2 } \end{aligned}$$

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

8 Functions f and g are defined by $$\begin{array} { l l } \mathrm { f } : x \mapsto 4 x - 2 k & \text { for } x \in \mathbb { R } , \text { where } k \text { is a constant, } \\ \mathrm { g } : x \mapsto \frac { 9 } { 2 - x } & \text { for } x \in \mathbb { R } , x \neq 2 . \end{array}$$

1. Find the values of \(k\) for which the equation \(\mathrm { fg } ( x ) = x\) has two equal roots.
2. Determine the roots of the equation \(\operatorname { fg } ( x ) = x\) for the values of \(k\) found in part (i).

10 The function f is defined by \(\mathrm { f } : x \mapsto 2 x + k , x \in \mathbb { R }\), where \(k\) is a constant.

1. In the case where \(k = 3\), solve the equation \(\mathrm { ff } ( x ) = 25\). The function g is defined by \(\mathrm { g } : x \mapsto x ^ { 2 } - 6 x + 8 , x \in \mathbb { R }\).
2. Find the set of values of \(k\) for which the equation \(\mathrm { f } ( x ) = \mathrm { g } ( x )\) has no real solutions. The function \(h\) is defined by \(h : x \mapsto x ^ { 2 } - 6 x + 8 , x > 3\).
3. Find an expression for \(\mathrm { h } ^ { - 1 } ( x )\).

1 Functions f and g are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto 10 - 3 x , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \mapsto \frac { 10 } { 3 - 2 x } , \quad x \in \mathbb { R } , x \neq \frac { 3 } { 2 } \end{aligned}$$ Solve the equation \(\mathrm { ff } ( x ) = \mathrm { gf } ( 2 )\).

6 The functions f and g are defined for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\) by $$\begin{aligned} & f ( x ) = \frac { 1 } { 2 } x + \frac { 1 } { 6 } \pi \\ & g ( x ) = \cos x \end{aligned}$$ Solve the following equations for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. \(\operatorname { gf } ( x ) = 1\), giving your answer in terms of \(\pi\).
2. \(\operatorname { fg } ( x ) = 1\), giving your answers correct to 2 decimal places.

6 The functions f and g are defined by $$\begin{aligned} & \mathrm { f } ( x ) = \frac { 2 } { x ^ { 2 } - 1 } \text { for } x < - 1 \\ & \mathrm {~g} ( x ) = x ^ { 2 } + 1 \text { for } x > 0 \end{aligned}$$

1. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
2. Solve the equation \(\operatorname { gf } ( x ) = 5\).

4. The functions f and g are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = \frac { 4 x + 6 } { x - 5 } & x \in \mathbb { R } , x \neq 5 \\ \mathrm {~g} ( x ) = 5 - 2 x ^ { 2 } & x \in \mathbb { R } , x \leqslant 0 \end{array}$$

1. Solve the equation $$\operatorname { fg } ( x ) = 3$$
2. Find \(\mathrm { f } ^ { - 1 }\)
3. Sketch and label, on the same axes, the curve with equation \(y = \mathrm { g } ( x )\) and the curve with equation \(y = \mathrm { g } ^ { - 1 } ( x )\). Show on your sketch the coordinates of the points where each curve meets or cuts the coordinate axes.

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 \(\mathrm { g } ( a ) = \mathrm { f } ^ { - 1 } ( a )\)

5. 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 \(g f ( - 1 )\).
3. Solve the equation $$\operatorname { fg } ( x ) = 17$$

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

6. 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.
Find, in terms of \(a\),

1. an expression for \(\mathrm { f } ^ { - 1 } ( x )\),
2. the range of g . Given that \(g f ( 3 ) = 7\),
3. find the two possible values of \(a\).

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

1.The function f is given by $$\mathrm { f } ( x ) = \sqrt { x + 2 } \quad \text { for } \quad x \in \mathbb { R } , x \geqslant 0$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\) The function g is given by $$\mathrm { g } ( x ) = x ^ { 2 } - 4 x + 5 \text { for } x \in \mathbb { R } , x \geqslant 0$$
2. Find the range of g .
3. Solve the equation \(\operatorname { fg } ( x ) = x\) .

7 The function f is defined for all real values of \(x\) by \(\mathrm { f } ( x ) = 2 x + 5\). The function g is defined for all real values of \(x\) and is such that \(\mathrm { g } ^ { - 1 } ( x ) = \sqrt [ 3 ] { x - a }\), where \(a\) is a constant. It is given that \(\mathrm { fg } ^ { - 1 } ( 12 ) = 9\). Find the value of \(a\) and hence solve the equation \(\operatorname { gf } ( x ) = 68\).

8 The functions \(f\) and \(g\) are defined as follows: $$\begin{gathered} \mathrm { f } ( x ) = 2 + \ln ( x + 3 ) \text { for } x \geqslant 0 \\ \mathrm {~g} ( x ) = a x ^ { 2 } \text { for all real values of } x , \text { where } a \text { is a positive constant. } \end{gathered}$$

1. Given that \(\operatorname { gf } \left( \mathrm { e } ^ { 4 } - 3 \right) = 9\), find the value of \(a\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\).
3. Given that \(\mathrm { ff } \left( \mathrm { e } ^ { N } - 3 \right) = \ln \left( 53 \mathrm { e } ^ { 2 } \right)\), find the value of \(N\).

4 In this question you must show detailed reasoning.
The functions f and g are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = x ^ { 3 } \quad \text { and } \quad \mathrm { g } ( x ) = x ^ { 2 } + 2 .$$

1. Write down expressions for
   1. \(\mathrm { fg } ( x )\),
   2. \(\operatorname { gf } ( x )\).
   3. Hence find the values of \(x\) for which \(\mathrm { fg } ( x ) - \mathrm { gf } ( x ) = 24\).

1. The functions f and g are defined by

$$\begin{array} { l l } f ( x ) = 4 - 3 x ^ { 2 } & x \in \mathbb { R } \\ g ( x ) = \frac { 5 } { 2 x - 9 } & x \in \mathbb { R } , x \neq \frac { 9 } { 2 } \end{array}$$

1. Find fg(2)
2. Find \(\mathrm { g } ^ { - 1 }\)
3. 1. Find \(\mathrm { gf } ( x )\), giving your answer as a simplified fraction.
   2. Deduce the range of \(\operatorname { gf } ( x )\). The function h is defined by $$h ( x ) = 2 x ^ { 2 } - 6 x + k \quad x \in \mathbb { R }$$ where \(k\) is a constant.
4. Find the range of values of \(k\) for which the equation $$\mathrm { f } ( x ) = \mathrm { h } ( x )$$ has no real solutions.

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

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 \(\operatorname { fg } ( x ) = 1\).

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 \(\mathrm { h } ( x ) = 3\).

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 \(\operatorname { gf } ( x ) = 6\).

1. 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 \(\operatorname { gf } ( x ) = 6\).

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 \(a\).