Find inverse function

3 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Write down the range of the function \(\mathrm { f } ( x )\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).

5 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\). The endpoints of the curve are \(\mathrm { P } ( - \pi , 1 )\) and \(\mathrm { Q } ( \pi , 3 )\), and \(\mathrm { f } ( x ) = a + \sin b x\), where \(a\) and \(b\) are constants. \begin{figure}[h] \end{figure}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac { 1 } { 2 }\).
2. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,2 ). Show that there is no point on the curve at which the gradient is greater than this.
3. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range. Write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 2,0 )\).
4. Find the area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\).

4 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 \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x \mathrm { r } } { 2 } \right)\) 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]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-2_499_562_779_785} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .

8 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Write down the range of the function \(\mathrm { f } ( x )\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).

5 Fig. 7 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + 2 \arctan x , x \in \mathbb { R }\). The scales on the \(x\) - and \(y\)-axes are the same. \begin{figure}[h] \end{figure}

1. Find the range of f , giving your answer in terms of \(\pi\).
2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and add a sketch of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) to the copy of Fig. 7.

3 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Write down the range of the function \(\mathrm { f } ( x )\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).

1. The figure 1 shows part of the graph of \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { a x + 4 } { x - b } , \quad x > 2\)

\begin{figure}[h] \end{figure} a. State the values of \(a\) and \(b\).
b. State the range of f.
c. Find \(\mathrm { f } ^ { - 1 } ( x )\), stating its domain.

9 The function \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 1 - \mathrm { e } ^ { x } }\) is defined on the domain \(x \in \mathbb { R } , x \neq 0\).

1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
2. Write down the range of \(\mathrm { f } ^ { - 1 } ( x )\).

4 The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = 3 \cos \frac { 1 } { 2 } x , & \text { for } 0 \leqslant x \leqslant 2 \pi \\ \mathrm {~g} ( x ) = | x | , & \text { for all real values of } x \end{array}$$

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 \(\mathrm { f } ^ { - 1 } ( x ) = 1\), giving your answer in an exact form.
3. 1. Write down an expression for \(\mathrm { gf } ( x )\).
   2. Sketch the graph of \(y = \operatorname { gf } ( x )\) for \(0 \leqslant x \leqslant 2 \pi\).
4. Describe a sequence of two geometrical transformations that maps the graph of \(y = \cos x\) onto the graph of \(y = 3 \cos \frac { 1 } { 2 } x\).

5 The function f is defined by $$\mathrm { f } ( x ) = \frac { x ^ { 2 } - 4 } { 3 } , \text { for real values of } x , \text { where } \boldsymbol { x } \leqslant \mathbf { 0 }$$

1. State the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Write down the domain of \(\mathrm { f } ^ { - 1 }\).
   2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. The function g is defined by $$\mathrm { g } ( x ) = \ln | 3 x - 1 | , \quad \text { for real values of } x , \text { where } x \neq \frac { 1 } { 3 }$$ The curve with equation \(y = \mathrm { g } ( x )\) is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{b8614dd6-2197-40c3-a673-5bef3e3653a5-6_469_819_1254_612}
   1. The curve \(y = \mathrm { g } ( x )\) intersects the \(x\)-axis at the origin and at the point \(P\). Find the \(x\)-coordinate of \(P\).
   2. State whether the function \(g\) has an inverse. Give a reason for your answer.
   3. Show that \(\operatorname { gf } ( x ) = \ln \left| x ^ { 2 } - k \right|\), stating the value of the constant \(k\).
   4. Solve the equation \(\mathrm { gf } ( x ) = 0\).

2 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = x - 2 & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 6 } { x + 3 } & \text { for real values of } x , \quad x \neq - 3 \end{array}$$ The composite function fg is denoted by h .

1. Find \(\mathrm { h } ( x )\).
2. 1. Find \(\mathrm { h } ^ { - 1 } ( x )\), where \(\mathrm { h } ^ { - 1 }\) is the inverse of h .
   2. Find the range of \(\mathrm { h } ^ { - 1 }\).

8 A function f is defined by \(\mathrm { f } ( x ) = 2 \mathrm { e } ^ { 3 x } - 1\) for all real values of \(x\).

1. Find the range of f.
2. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \frac { 1 } { 3 } \ln \left( \frac { x + 1 } { 2 } \right)\).
3. Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) when \(x = 0\).

1. The function f is defined by

$$\mathrm { f } ( x ) \equiv 2 + \ln ( 3 x - 2 ) , \quad x \in \mathbb { R } , \quad x > \frac { 2 } { 3 } .$$

1. Find the exact value of \(\mathrm { ff } ( 1 )\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

7. \(\quad f ( x ) = x ^ { 2 } - 2 x + 5 , x \in \mathbb { R } , x \geq 1\).

1. Express \(\mathrm { f } ( x )\) in the form \(( x + a ) ^ { 2 } + b\), where \(a\) and \(b\) are constants.
2. State the range of f.
3. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
4. Describe fully two transformations that would map the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\) onto the graph of \(y = \sqrt { x } , x \geq 0\).
5. Find an equation for the normal to the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point where \(x = 8\).

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

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

6 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } - 3 , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 1 } { 3 x + 4 } , & \text { for real values of } x , x \neq - \frac { 4 } { 3 } \end{array}$$

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 \(\mathrm { f } ^ { - 1 } ( x ) = 0\).
3. 1. Find an expression for \(\operatorname { gf } ( x )\).
   2. Solve the equation \(\mathrm { gf } ( x ) = 1\), giving your answer in an exact form.

4 The functions f and g are defined by $$\begin{array} { l l } \mathrm { f } ( x ) = 5 - \mathrm { e } ^ { 3 x } , & \text { for all real values of } x \\ \mathrm {~g} ( x ) = \frac { 1 } { 2 x - 3 } , & \text { for } x \neq 1.5 \end{array}$$

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 \(\mathrm { f } ^ { - 1 } ( x ) = 0\).
3. Find an expression for \(\operatorname { gg } ( x )\), giving your answer in the form \(\frac { a x + b } { c x + d }\), where \(a , b , c\) and \(d\) are integers.
   [0pt] [3 marks]

4 The function f is defined by \(\mathrm { f } ( x ) = \mathrm { e } ^ { x - 4 } , x \in \mathbb { R }\) Find \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain.

13 The function f is defined by $$\mathrm { f } ( x ) = \frac { 2 x + 3 } { x - 2 } \quad x \in \mathbb { R } , x \neq 2$$ 13

1. 1. Find f-1
      13
      1. (ii) Write down an expression for \(\mathrm { ff } ( x )\).
         13
   2. The function g is defined by $$g ( x ) = \frac { 2 x ^ { 2 } - 5 x } { 2 } \quad x \in \mathbb { R } , 0 \leq x \leq 4$$ 13
   3. 1. Find the range of g .
         13
   4. (ii) Determine whether g has an inverse.
      Fully justify your answer.
   5. Show that $$g f ( x ) = \frac { 48 + 29 x - 2 x ^ { 2 } } { 2 x ^ { 2 } - 8 x + 8 }$$ 13
   6. It can be shown that fg is given by $$f g ( x ) = \frac { 4 x ^ { 2 } - 10 x + 6 } { 2 x ^ { 2 } - 5 x - 4 }$$ with domain \(\{ x \in \mathbb { R } : 0 \leq x \leq 4 , x \neq a \}\) Find the value of \(a\).
      Fully justify your answer.

7 The functions f and g are defined by $$\begin{aligned} & \mathrm { f } ( x ) = \sqrt { 10 - 2 x } \text { for } \quad x \leq 5 \\ & \mathrm {~g} ( x ) = \frac { 1 } { x } \quad \text { for } \quad x \neq 0 \end{aligned}$$ The function \(h\) has maximum possible domain and is defined by $$\mathrm { h } ( x ) = \operatorname { gf } ( x )$$ 7

1. Find an expression for \(\mathrm { h } ( x )\) 7
2. Find the domain of h
   7
3. Show that \(\mathrm { h } ^ { - 1 } ( x ) = 5 - \frac { 1 } { 2 x ^ { 2 } }\) \includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-11_2488_1716_219_153}

1 The function \(\mathrm { f } ( x )\) is defined for all real \(x\) by \(f ( x ) = 3 x - 2\).

1. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
2. Sketch the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) on the same diagram.
3. Find the set of values of \(x\) for which \(\mathrm { f } ( x ) > \mathrm { f } ^ { - 1 } ( x )\).

2 The function f is defined by \(\mathrm { f } ( x ) = \frac { 2 x + 1 } { x - 3 }\) for all real \(x , x \neq 3\). Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\).

\includegraphics{figure_6} The function f is defined by f\((x) = \frac{2}{x^2} + 4\) for \(x < 0\). The diagram shows the graph of \(y = \text{f}(x)\).

1. On this diagram, sketch the graph of \(y = \text{f}^{-1}(x)\). Show any relevant mirror line. [2]
2. Find an expression for f\(^{-1}(x)\). [3]
3. Solve the equation f\((x) = 4.5\). [1]
4. Explain why the equation f\(^{-1}(x) = \text{f}(x)\) has no solution. [1]

The function f is defined as follows: $$f(x) = \sqrt{x-1} \text{ for } x > 1.$$ \begin{enumerate}[label=(\alph*)] \item Find an expression for \(f^{-1}(x)\). [1] \end enumerate} \includegraphics{figure_4} The diagram shows the graph of \(y = g(x)\) where \(g(x) = \frac{1}{x^2+2}\) for \(x \in \mathbb{R}\). \begin{enumerate}[label=(\alph*)] \setcounter{enumi}{1} \item State the range of g and explain whether \(g^{-1}\) exists. [2] \end enumerate} The function h is defined by \(h(x) = \frac{1}{x^2+2}\) for \(x \geqslant 0\). \begin{enumerate}[label=(\alph*)] \setcounter{enumi}{2} \item Solve the equation \(hf(x) = f\left(\frac{25}{16}\right)\). Give your answer in the form \(a + b\sqrt{c}\), where \(a\), \(b\) and \(c\) are integers. [4] \end enumerate}

Functions f and g are defined by $$f(x) = (x + a)^2 - a \text{ for } x \leqslant -a,$$ $$g(x) = 2x - 1 \text{ for } x \in \mathbb{R},$$ where \(a\) is a positive constant.

1. Find an expression for \(f^{-1}(x)\). [3]
2. 1. State the domain of the function \(f^{-1}\). [1]
   2. State the range of the function \(f^{-1}\). [1]
3. Given that \(a = \frac{7}{2}\), solve the equation \(gf(x) = 0\). [3]