Solve equation involving composites

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 $$\mathrm { fg } ^ { - 1 } ( x ) = 2$$

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}

7 The function f is defined by $$\mathrm { f } ( x ) = \frac { a x - 5 } { 2 x + b } \quad x \in \mathbb { R } , x \neq \frac { 9 } { 2 }$$ where \(a\) and \(b\) are integers.
The graph of \(y = \mathrm { f } ( x )\) has asymptotes \(x = \frac { 9 } { 2 }\) and \(y = 3\) 7

1. Find the value of \(a\) and the value of \(b\) 7
2. Solve the inequality $$\mathrm { f } ( x ) \leq x + 2$$ Fully justify your answer.

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

1. Show that \(f(x) = \frac{1}{(x-1)(x+1)}\). [3]
2. Find the range of \(f\). [2]

The function \(g\) is given by $$g : x \mapsto \frac{2}{x}, \quad x > 0.$$

1. Solve \(gf(x) = 70\). [4]

The function f is given by $$f : x \alpha \frac{x}{x^2-1} - \frac{1}{x+1}, \quad x > 1.$$

1. Show that \(\text{f}(x) = \frac{1}{(x-1)(x+1)}\). [3]
2. Find the range of f. [2]

The function g is given by $$g : x \alpha \frac{2}{x}, \quad x > 0.$$

1. Solve gf(x) = 70. [4]

The functions \(f(x)\) and \(g(x)\) are defined by \(f(x) = \ln x\) and \(g(x) = 2 + e^x\), for \(x > 0\). Find the exact value of \(x\), given that \(fg(x) = 2x\). [5]

\includegraphics{figure_5} The functions f(x) and g(x) are defined for \(x \geqslant 0\) by \(\text{f}(x) = \frac{x}{x^2 + 3}\) and \(\text{g}(x) = \text{e}^{-2x}\). The diagram shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The equation \(\text{f}(x) = \text{g}(x)\) has exactly one real root \(\alpha\).

1. Show that \(\alpha\) satisfies the equation \(\text{h}(x) = 0\), where \(\text{h}(x) = x^2 + 3 - x\text{e}^{2x}\). [2]
2. Hence show that a Newton-Raphson iterative formula for finding \(\alpha\) can be written in the form $$x_{n+1} = \frac{x_n^2(1 - 2\text{e}^{2x_n}) - 3}{2x_n - (1 + 2x_n)\text{e}^{2x_n}}.$$ [5]
3. Use this iterative formula, with a suitable initial value, to find \(\alpha\) correct to 3 decimal places. Show the result of each iteration. [3]
4. In this question you must show detailed reasoning. Find the exact value of \(x\) for which \(\text{fg}(x) = \frac{2}{13}\). [6]