1.02v Inverse and composite functions: graphs and conditions for existence

1. The function f is defined by

$$f ( x ) = \frac { 3 x - 7 } { x - 2 } \quad x \in \mathbb { R } , x \neq 2$$

1. Find \(f ^ { - 1 } ( 7 )\)
2. Show that \(\operatorname { ff } ( x ) = \frac { a x + b } { x - 3 }\) where \(a\) and \(b\) are integers to be found.

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1. The function f is defined by

$$f : x \mapsto \frac { 3 x - 5 } { x + 1 } , \quad x \in \mathbb { R } , \quad x \neq - 1$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
2. Show that $$\mathrm { ff } ( x ) = \frac { x + a } { x - 1 } , \quad x \in \mathbb { R } , \quad x \neq \pm 1$$ where \(a\) is an integer to be found. The function g is defined by $$\mathrm { g } : x \mapsto x ^ { 2 } - 3 x , \quad x \in \mathbb { R } , 0 \leqslant x \leqslant 5$$
3. Find the value of \(\mathrm { fg } ( 2 )\).
4. Find the range of g.
5. Explain why the function \(g\) does not have an inverse.

1. $$\operatorname { g } ( x ) = \frac { 2 x + 5 } { x - 3 } \quad x \geqslant 5$$

1. Find \(\mathrm { gg } ( 5 )\).
2. State the range of g.
3. Find \(\mathrm { g } ^ { - 1 } ( x )\), stating its domain.

6. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the graph of \(y = \mathrm { g } ( x )\), where $$g ( x ) = \begin{cases} ( x - 2 ) ^ { 2 } + 1 & x \leqslant 2 \\ 4 x - 7 & x > 2 \end{cases}$$

1. Find the value of \(\operatorname { gg } ( 0 )\).
2. Find all values of \(x\) for which $$\mathrm { g } ( x ) > 28$$ The function h is defined by $$\mathrm { h } ( x ) = ( x - 2 ) ^ { 2 } + 1 \quad x \leqslant 2$$
3. Explain why h has an inverse but g does not.
4. Solve the equation $$\mathrm { h } ^ { - 1 } ( x ) = - \frac { 1 } { 2 }$$

1. The function f is defined by

$$f ( x ) = \frac { 8 x + 5 } { 2 x + 3 } \quad x > - \frac { 3 } { 2 }$$

1. Find \(\mathrm { f } ^ { - 1 } \left( \frac { 3 } { 2 } \right)\)
2. Show that $$\mathrm { f } ( x ) = A + \frac { B } { 2 x + 3 }$$ where \(A\) and \(B\) are constants to be found. The function \(g\) is defined by $$g ( x ) = 16 - x ^ { 2 } \quad 0 \leqslant x \leqslant 4$$
3. State the range of \(\mathrm { g } ^ { - 1 }\)
4. Find the range of \(\mathrm { fg } ^ { - 1 }\)

1. The function \(g\) is defined by

$$g ( x ) = \frac { 3 \ln ( x ) - 7 } { \ln ( x ) - 2 } \quad x > 0 \quad x \neq k$$ where \(k\) is a constant.

1. Deduce the value of \(k\).
2. Prove that $$\mathrm { g } ^ { \prime } ( x ) > 0$$ for all values of \(x\) in the domain of g .
3. Find the range of values of \(a\) for which $$g ( a ) > 0$$

1. The functions f and g are defined by

$$\begin{aligned} & f ( x ) = 7 - 2 x ^ { 2 } \quad x \in \mathbb { R } \\ & \operatorname { g } ( x ) = \frac { 3 x } { 5 x - 1 } \quad x \in \mathbb { R } \quad x \neq \frac { 1 } { 5 } \end{aligned}$$

1. State the range of f
2. Find gf (1.8)
3. Find \(\mathrm { g } ^ { - 1 } ( x )\)

5

1. (A) Sketch the graph of \(y = 3 ^ { x }\).
   (B) Give the coordinates of any intercepts. The curve \(y = \mathrm { f } ( x )\) is the reflection of the curve \(y = 3 ^ { x }\) in the line \(y = x\).
2. Find \(\mathrm { f } ( x )\).

9 You are given that \(\mathrm { f } ( x ) = 2 x + 3 \quad\) for \(x < 0 \quad\) and \(\mathrm { g } ( x ) = x ^ { 2 } - 2 x + 1\) for \(x > 1\).

1. Find \(\mathrm { gf } ( x )\), stating the domain.
2. State the range of \(\mathrm { gf } ( x )\).
3. Find (gf) \({ } ^ { - 1 } ( x )\).

12 It is given that

• \(\mathrm { f } ( x ) = \pm \frac { 1 } { \sqrt { x } } , x > 0\)
• \(\mathrm { g } ( x ) = \frac { x } { x - 3 } , x > 3\)
• \(\mathrm { h } ( x ) = x ^ { 2 } + 2 , x \in \mathbb { R }\).
  1. Explain why \(\mathrm { f } ( x )\) is not a function.
  2. Find \(\mathrm { gh } ( x )\).
  3. State the domain of \(\mathrm { gh } ( x )\).

15 Functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined as follows. \(\mathrm { f } ( x ) = \sqrt { x }\) for \(x > 0\) and \(\mathrm { g } ( x ) = x ^ { 3 } - x - 6\) for \(x > 2\). The function \(\mathrm { h } ( x )\) is defined as \(\mathrm { h } ( x ) = \mathrm { fg } ( x )\).

1. Find \(\mathrm { h } ( x )\) in terms of \(x\) and state its domain.
2. Find \(\mathrm { h } ( 3 )\). Fig. 15 shows \(\mathrm { h } ( x )\) and \(\mathrm { h } ^ { - 1 } ( x )\), together with the straight line \(y = x\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{cea67565-8074-4703-8e1a-09b98e380baf-17_780_796_895_242} \captionsetup{labelformat=empty} \caption{Fig. 15}
   \end{figure}
3. Determine the gradient of \(\mathrm { y } = \mathrm { h } ^ { - 1 } ( \mathrm { x } )\) at the point where \(y = 3\).

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 \(\mathrm { f } ^ { - 1 } ( x )\).

8 The curves \(\mathrm { y } = \mathrm { h } ( \mathrm { x } )\) and \(\mathrm { y } = \mathrm { h } ^ { - 1 } ( \mathrm { x } )\), where \(\mathrm { h } ( x ) = x ^ { 3 } - 8\), are shown below.
The curve \(\mathrm { y } = \mathrm { h } ( \mathrm { x } )\) crosses the \(x\)-axis at B and the \(y\)-axis at A.
The curve \(\mathrm { y } = \mathrm { h } ^ { - 1 } ( \mathrm { x } )\) crosses the \(x\)-axis at D and the \(y\)-axis at C . \includegraphics[max width=\textwidth, alt={}, center]{c30a926b-d832-46f5-aa65-0066ef482c3d-7_826_819_520_255}

1. Find an expression for \(\mathrm { h } ^ { - 1 } ( x )\).
2. Determine the coordinates of A, B, C and D.
3. Determine the equation of the perpendicular bisector of AB . Give your answer in the form \(\mathrm { y } = \mathrm { mx } + c\), where \(m\) and \(c\) are constants to be determined.
4. Points A , B , C and D lie on a circle. Determine the equation of the circle. Give your answer in the form \(( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } = r ^ { 2 }\), where \(a\), \(b\) and \(r ^ { 2 }\) are constants to be determined.

2

1. The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = \sqrt { 1 + 2 x } \text { for } x \geqslant - \frac { 1 } { 2 }$$ Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of this inverse function.
2. Explain why \(\mathrm { g } ( x ) = 1 + x ^ { 2 }\), with domain all real numbers, has no inverse function.

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

1. Find the range of f.
2. The inverse of g is \(\mathrm { g } ^ { - 1 }\).
   1. Find \(\mathrm { g } ^ { - 1 } ( x )\).
   2. State the range of \(\mathrm { g } ^ { - 1 }\).
3. The composite function gf is denoted by h .
   1. Find \(\mathrm { h } ( x )\), simplifying your answer.
   2. State the greatest possible domain of h .

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

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 \(y = 3 \cos \frac { 1 } { 2 } x\).

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

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 \(x = 0\).

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

3

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when:
   1. \(\quad y = \ln ( 5 x - 2 )\);
   2. \(y = \sin 2 x\).
2. The functions f and g are defined with their respective domains by $$\begin{array} { l l } \mathrm { f } ( x ) = \ln ( 5 x - 2 ) , & \text { for real values of } x \text { such that } x \geqslant \frac { 1 } { 2 } \\ \mathrm {~g} ( x ) = \sin 2 x , & \text { for real values of } x \text { in the interval } - \frac { \pi } { 4 } \leqslant x \leqslant \frac { \pi } { 4 } \end{array}$$
   1. Find the range of f .
   2. Find an expression for \(\operatorname { gf } ( x )\).
   3. Solve the equation \(\operatorname { gf } ( x ) = 0\).
   4. The inverse of g is \(\mathrm { g } ^ { - 1 }\). Find \(\mathrm { g } ^ { - 1 } ( x )\).

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 \(\mathrm { f } ^ { - 1 } ( x ) = 7\).