Determine if inverse exists

9 Functions f and g are defined by $$\begin{aligned} & \mathrm { f } ( x ) = x + \frac { 1 } { x } \quad \text { for } x > 0 \\ & \mathrm {~g} ( x ) = a x + 1 \quad \text { for } x \in \mathbb { R } \end{aligned}$$ where \(a\) is a constant.

1. Find an expression for \(\operatorname { gf } ( x )\).
2. Given that \(\operatorname { gf } ( 2 ) = 11\), find the value of \(a\).
3. Given that the graph of \(y = \mathrm { f } ( x )\) has a minimum point when \(x = 1\), explain whether or not f has an inverse.
   It is given instead that \(a = 5\).
4. Find and simplify an expression for \(\mathrm { g } ^ { - 1 } \mathrm { f } ( x )\).
5. Explain why the composite function fg cannot be formed. \includegraphics[max width=\textwidth, alt={}, center]{5d26c357-ea9f-47d9-8eca-2152901cf2f1-16_1143_1008_267_566} The diagram shows a cross-section RASB of the body of an aircraft. The cross-section consists of a sector \(O A R B\) of a circle of radius 2.5 m , with centre \(O\), a sector \(P A S B\) of another circle of radius 2.24 m with centre \(P\) and a quadrilateral \(O A P B\). Angle \(A O B = \frac { 2 } { 3 } \pi\) and angle \(A P B = \frac { 5 } { 6 } \pi\).

10 The functions \(f\) and \(g\) are defined as follows: $$\begin{array} { l l } \mathrm { f } : x \mapsto x ^ { 2 } - 2 x , & x \in \mathbb { R } , \\ \mathrm {~g} : x \mapsto 2 x + 3 , & x \in \mathbb { R } . \end{array}$$

1. Find the set of values of \(x\) for which \(\mathrm { f } ( x ) > 15\).
2. Find the range of f and state, with a reason, whether f has an inverse.
3. Show that the equation \(\operatorname { gf } ( x ) = 0\) has no real solutions.
4. Sketch, in a single diagram, the graphs of \(y = \mathrm { g } ( x )\) and \(y = \mathrm { g } ^ { - 1 } ( x )\), making clear the relationship between the graphs.

11 The function f is defined by \(\mathrm { f } : x \mapsto 2 x ^ { 2 } - 8 x + 11\) for \(x \in \mathbb { R }\).

1. Express \(\mathrm { f } ( x )\) in the form \(a ( x + b ) ^ { 2 } + c\), where \(a\), \(b\) and \(c\) are constants.
2. State the range of f .
3. Explain why f does not have an inverse. The function g is defined by \(\mathrm { g } : x \mapsto 2 x ^ { 2 } - 8 x + 11\) for \(x \leqslant A\), where \(A\) is a constant.
4. State the largest value of \(A\) for which g has an inverse.
5. When \(A\) has this value, obtain an expression, in terms of \(x\), for \(\mathrm { g } ^ { - 1 } ( x )\) and state the range of \(\mathrm { g } ^ { - 1 }\).

3

1. State the algebraic condition for the function \(\mathrm { f } ( x )\) to be an even function.
   What geometrical property does the graph of an even function have?
2. State whether the following functions are odd, even or neither.
   (A) \(\mathrm { f } ( x ) = x ^ { 2 } - 3\) (B) \(\mathrm { g } ( x ) = \sin x + \cos x\) (C) \(\mathrm { h } ( x ) = \frac { 1 } { x + x ^ { 3 } }\)

3 The function \(f ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h] \end{figure} (1) Show algebraically that the function is even. State how this property relates to the shape of the curve.
(ii) Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
(iii) Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\),
(iv) Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(g ( x )\).
Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
(v) Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the commection between this result and your answer to part (ii).

7 The function f is defined for \(x > 0\) by \(\mathrm { f } ( x ) = \ln x\) and the function g is defined for all real values of \(x\) by \(\mathrm { g } ( x ) = x ^ { 2 } + 8\).

1. Find the exact, positive value of \(x\) which satisfies the equation \(\operatorname { fg } ( x ) = 8\).
2. State which one of f and g has an inverse and define that inverse function.
3. Find the exact value of the gradient of the curve \(y = \operatorname { gf } ( x )\) at the point with \(x\)-coordinate \(\mathrm { e } ^ { 3 }\).
4. Use Simpson's rule with four strips to find an approximate value of $$\int _ { - 4 } ^ { 4 } \mathrm { fg } ( x ) \mathrm { d } x$$ giving your answer correct to 3 significant figures.

5. The function f is defined by $$\mathrm { f } : x \rightarrow \frac { 2 x - 3 } { x - 1 } \quad x \in R , x \neq 1$$ a. Find \(f ^ { - 1 } ( 3 )\).
b. Show that $$\mathrm { ff } ( x ) = \frac { x + p } { x - 2 } \quad x \in R , \quad x \neq 2$$ where \(p\) is an integer to be found. The function g is defined by $$g : x \rightarrow x ^ { 2 } - 5 x \quad x \in R , 0 \leq x \leq 6$$ c. Find the range of g .
d. Explain why the function g does not have an inverse.

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.

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

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 \(\mathrm { gh } ( x )\).

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.

10 The function h is defined by $$\mathrm { h } ( x ) = \frac { \sqrt { x } } { x - 3 }$$ where \(h\) has its maximum possible domain.
10

1. Find the domain of h .
   Give your answer using set notation. 10
2. Alice correctly calculates $$h ( 1 ) = - 0.5 \text { and } h ( 4 ) = 2$$ She then argues that since there is a change of sign there must be a value of \(x\) in the interval \(1 < x < 4\) that gives \(\mathrm { h } ( x ) = 0\) Explain the error in Alice's argument.
   [0pt] [2 marks]
   10
3. By considering any turning points of h , determine whether h has an inverse function. Fully justify your answer.
   [0pt] [6 marks]

5 Let \(\mathrm { f } ( x ) = x ^ { 2 }\) and \(\mathrm { g } ( x ) = 7 x - 2\) for all real values of \(x\).

1. Give a reason why f has no inverse function.
2. Write down an expression for \(\mathrm { gf } ( x )\).
3. Find \(\mathrm { g } ^ { - 1 } ( x )\).

11 The functions f and g are defined by \(\mathrm { f } ( x ) = \frac { 1 } { 2 + x } + 5 , x > - 2\) and \(\mathrm { g } ( x ) = | x | , x \in \mathbb { R }\).

1. Given that the range of f is of the form \(\mathrm { f } ( x ) > a\), find \(a\).
2. Find an expression for \(\mathrm { f } ^ { - 1 }\), stating its domain and range.
3. Show that \(\mathrm { gf } ( x ) = \mathrm { f } ( x )\).
4. Find an expression for \(\mathrm { fg } ( x )\). Determine whether fg has an inverse.

3 Let \(\mathrm { f } ( x ) = x ^ { 2 }\) and \(\mathrm { g } ( x ) = 7 x - 2\) for all real values of \(x\).

1. Give a reason why f has no inverse function.
2. Write down an expression for \(\operatorname { gf } ( x )\).
3. Find \(\mathrm { g } ^ { - 1 } ( x )\).
4. Explain the relationship between the graph of \(y = \mathrm { g } ( x )\) and \(y = \mathrm { g } ^ { - 1 } ( x )\).

3 Let \(\mathrm { f } ( x ) = x ^ { 2 }\) and \(\mathrm { g } ( x ) = 7 x - 2\) for all real values of \(x\).

1. Give a reason why f has no inverse function.
2. Write down an expression for \(\operatorname { gf } ( x )\).
3. Find \(\mathrm { g } ^ { - 1 } ( x )\).
4. Explain the relationship between the graph of \(y = \mathrm { g } ( x )\) and \(y = \mathrm { g } ^ { - 1 } ( x )\).

The functions f and g are defined with their respective domains by $$f(x) = x^2 \quad \text{for all real values of } x$$ $$g(x) = \frac{1}{2x + 1} \quad \text{for real values of } x, \quad x \neq -0.5$$

1. Explain why f does not have an inverse. [1]
2. The inverse of g is \(g^{-1}\). Find \(g^{-1}(x)\). [3]
3. State the range of \(g^{-1}\). [1]
4. Solve the equation \(fg(x) = g(x)\). [3]

The functions f and g are defined for all real values of \(x\) by $$f(x) = 4x^2 - 12x \quad \text{and} \quad g(x) = ax + b,$$ where \(a\) and \(b\) are non-zero constants.

1. Find the range of f. [3]
2. Explain why the function f has no inverse. [2]
3. Given that \(g^{-1}(x) = g(x)\) for all values of \(x\), show that \(a = -1\). [4]
4. Given further that gf\((x) < 5\) for all values of \(x\), find the set of possible values of \(b\). [4]

The function \(f(x) = \ln(1 + x^2)\) has domain \(-3 \leq x \leq 3\). Fig. 9 shows the graph of \(y = f(x)\). \includegraphics{figure_9}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve. [3]
2. Find the gradient of the curve at the point P\((2, \ln 5)\). [4]
3. Explain why the function does not have an inverse for the domain \(-3 \leq x \leq 3\). [1]

The domain of \(f(x)\) is now restricted to \(0 \leq x \leq 3\). The inverse of \(f(x)\) is the function \(g(x)\).

1. Sketch the curves \(y = f(x)\) and \(y = g(x)\) on the same axes. State the domain of the function \(g(x)\). Show that \(g(x) = \sqrt{e^x - 1}\). [6]
2. Differentiate \(g(x)\). Hence verify that \(g'(\ln 5) = \frac{1}{4}\). Explain the connection between this result and your answer to part (ii). [5]

The function f is defined by $$f(x) = |x| + 1 \quad \text{for } x \in \mathbb{R}$$ The function g is defined by $$g(x) = \ln x$$ where g has its greatest possible domain.

1. Using set notation, state the range of f [2 marks]
2. State the domain of g [1 mark]
3. The composite function h is given by $$h(x) = g f(x) \quad \text{for } x \in \mathbb{R}$$
   1. Write down an expression for \(h(x)\) in terms of \(x\) [1 mark]
   2. Determine if h has an inverse. Fully justify your answer. [2 marks]

Each of these functions has domain \(x \in \mathbb{R}\) Which function does not have an inverse? Circle your answer. [1 mark] \(f(x) = x^3\) \quad \(f(x) = 2x + 1\) \quad \(f(x) = x^2\) \quad \(f(x) = e^x\)

Determine which one of these graphs does not represent \(y\) as a function of \(x\). Tick (\(\checkmark\)) one box. [1 mark] \includegraphics{figure_3}

Functions f, g and h are defined for \(x \in \mathbb{R}\) by $$f : x \mapsto x^2 - 2x,$$ $$g : x \mapsto x^2,$$ $$h : x \mapsto \sin x.$$

1. 1. State whether or not f has an inverse, giving a reason. [2]
   2. Determine the range of the function f. [2]
2. 1. Show that gh(x) can be expressed as \(\frac{1}{2}(1 - \cos 2x)\). [2]
   2. Sketch the curve C defined by \(y = \text{gh}(x)\) for \(0 \leqslant x \leqslant 2\pi\). [3]