Determine domain for composite

8 The function f is defined by \(\mathrm { f } ( x ) = 3 x + 1\) for \(x \leqslant a\), where \(a\) is a constant. The function g is defined by \(\mathrm { g } ( x ) = - 1 - x ^ { 2 }\) for \(x \leqslant - 1\).

1. Find the largest value of \(a\) for which the composite function gf can be formed. For the case where \(a = - 1\),
2. solve the equation \(\operatorname { fg } ( x ) + 14 = 0\),
3. find the set of values of \(x\) which satisfy the inequality \(\operatorname { gf } ( x ) \leqslant - 50\).

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.

4 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }\).
The curve has asymptotes \(x = 0\) and \(x = a\). \begin{figure}[h] \end{figure}

1. Find \(a\). Hence write down the domain of the function.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the coordinates of the turning point of the curve, and write down the range of the function. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
3. (A) Show algebraically that \(\mathrm { g } ( x )\) is an even function.
   (B) Show that \(\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )\).
   (C) Hence prove that the curve \(y = \mathrm { f } ( x )\) is symmetrical, and state its line of symmetry.

2.The functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = 2 \sqrt { 1 - \mathrm { e } ^ { - x } } & x \in \mathbb { R } , x \geqslant 0 \\ \mathrm {~g} ( x ) = \ln \left( 4 - x ^ { 2 } \right) & x \in \mathbb { R } , - 2 < x < 2 \end{array}$$

1. 1. Explain why fg cannot be formed as a composite function.
   2. Explain why gf can be formed as a composite function.
2. 1. Find \(\mathrm { gf } ( x )\) ,giving the answer in the form \(\mathrm { gf } ( x ) = a + b x\) ,where \(a\) and \(b\) are constants.
   2. State the domain and range of gf.
3. Sketch the graph of the function gf.
   On your sketch,you should show the coordinates of any points where the graph meets or crosses the coordinate axes. The circle \(C\) with centre \(( 0 , - \ln 9 )\) touches the line with equation \(y = \operatorname { gf } ( x )\) at precisely one point.
4. Find an equation of the circle \(C\) .

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }\).
The curve has asymptotes \(x = 0\) and \(x = a\). \begin{figure}[h] \end{figure}

1. Find \(a\). Hence write down the domain of the function.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the coordinates of the turning point of the curve, and write down the range of the function. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
3. (A) Show algebraically that \(\mathrm { g } ( x )\) is an even function.
   (B) Show that \(\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )\).
   (C) Hence prove that the curve \(y = \mathrm { f } ( x )\) is symmetrical, and state its line of symmetry.

2 The function \(\mathrm { f } ( x ) = \sqrt { x }\) is defined on the domain \(x \geqslant 0\).
The function \(\mathrm { g } ( x ) = 25 - x ^ { 2 }\) is defined on the domain \(\mathbb { R }\).

1. Write down an expression for \(\mathrm { fg } ( x )\).
2. 1. Find the domain of \(\mathrm { fg } ( x )\).
   2. Find the range of \(\mathrm { fg } ( x )\).

Functions f and g are defined by $$f(x) = 2 \sin x \quad \text{for } -\frac{1}{2}\pi \leq x \leq \frac{1}{2}\pi,$$ $$g(x) = 4 - 2x^2 \quad \text{for } x \in \mathbb{R}.$$

1. State the range of f and the range of g. [2]
2. Show that gf(0.5) = 2.16, correct to 3 significant figures, and explain why fg(0.5) is not defined. [4]
3. Find the set of values of \(x\) for which \(f^{-1}g(x)\) is not defined. [6]