1.02u Functions: definition and vocabulary (domain, range, mapping)

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) \(\mathrm { h } ( x ) = \frac { 1 } { x + x ^ { 3 } }\)

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { 2 x } } { 1 + \mathrm { e } ^ { 2 x } }\). The curve crosses the \(y\)-axis at P. \begin{figure}[h] \end{figure}

1. Find the coordinates of P .
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), simplifying your answer. Hence calculate the gradient of the curve at P .
3. Show that the area of the region enclosed by \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is \(\frac { 1 } { 2 } \ln \left( \frac { 1 + \mathrm { e } ^ { 2 } } { 2 } \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { 2 } \left( \frac { \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } } { \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } } \right)\).
4. Prove algebraically that \(\mathrm { g } ( x )\) is an odd function. Interpret this result graphically.
5. (A) Show that \(\mathrm { g } ( x ) + \frac { 1 } { 2 } = \mathrm { f } ( x )\).
   (B) Describe the transformation which maps the curve \(y = \mathrm { g } ( x )\) onto the curve \(y = \mathrm { f } ( x )\).
   (C) What can you conclude about the symmetry of the curve \(y = \mathrm { f } ( x )\) ?

8 You are given that \(\mathrm { f } ( x ) = \frac { x } { x ^ { 2 } + 1 }\) for all real values of \(x\).

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 1 - x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\).
2. Hence show that there is a stationary value at \(\left( 1 , \frac { 1 } { 2 } \right)\) and find the coordinates of the other stationary point.
3. The graph of the curve is shown in Fig. 8. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-3_518_892_1612_705} \captionsetup{labelformat=empty} \caption{Fig. 8}
   \end{figure} State whether the curve is odd or even and prove the result algebraically.
4. Show that \(\int _ { 1 } ^ { 4 } \frac { x } { x ^ { 2 } + 1 } \mathrm {~d} x = \int _ { a } ^ { b } k \frac { 1 } { u + 1 } \mathrm {~d} u\), where the values of \(a , b\) and \(k\) are to be determined.
5. Hence find the area of the shaded region in Fig. 8.

7 The functions \(f , g\) and \(h\) are defined as follows. $$\mathrm { f } ( x ) = 2 x \quad \mathrm {~g} ( x ) = x ^ { 2 } \quad \mathrm {~h} ( x ) = x + 2$$ Find each of the following as functions of \(x\).

1. \(\mathrm { f } ^ { 2 } ( x )\),
2. \(\operatorname { fgh } ( x )\),
3. \(\mathrm { h } ^ { - 1 } ( x )\).

2

1. Show that \(\mathrm { f } ( x ) = \left| x ^ { 3 } \right|\) is an even function.
2. It is suggested that the function \(\mathrm { g } ( x ) = ( x - 1 ) ^ { 3 }\) is odd. Prove that this is false.

3. 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 \(\mathrm { gf } ( 1 )\).
2. Find an expression for \(\mathrm { g } ^ { - 1 } ( x )\).
3. Find, in terms of natural logarithms, the solution of the equation $$\mathrm { fg } ^ { - 1 } ( x ) = 2$$

4 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
P is the point on the curve with \(x\)-coordinate \(\frac { 1 } { 3 } \pi\). \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P .
2. Find \(\mathrm { f } ^ { \prime } ( x )\). Hence find the gradient of the curve at the point P .
3. Show that the derivative of \(\frac { \sin x } { 1 + \cos x }\) is \(\frac { 1 } { 1 + \cos x }\). Hence find the exact area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 3 } \pi\).
4. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)\). State the domain of this inverse function, and add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.

1 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h] \end{figure}

1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by $$\sqrt { } \text { for } x \geqslant 0 \text {. }$$
2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
3. Show that \(\int \left( \mathrm { e } ^ { x } 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } \quad 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } \quad 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).

2 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x\). \begin{figure}[h] \end{figure}

1. Find the range of the function \(\mathrm { f } ( x )\), giving your answer in terms of \(\pi\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the origin.
3. Hence write down the gradient of \(y = \frac { 1 } { 2 } \arctan x\) at the origin.

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).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x _ { 2 } } }\). \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. (A) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

1 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 1 } { x ^ { 2 } + 1 }\) for the domain \(0 \leqslant x \leqslant 2\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\), and hence that \(\mathrm { f } ( x )\) is an increasing function for \(x > 0\).
2. Find the range of \(\mathrm { f } ( x )\).
3. Given that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 6 - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 3 } }\), find the maximum value of \(\mathrm { f } ^ { \prime } ( x )\). The function \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
4. Write down the domain and range of \(\mathrm { g } ( x )\). Add a sketch of the curve \(y = \mathrm { g } ( x )\) to a copy of Fig. 9 .
5. Show that \(\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }\).

6 Given that \(\mathrm { f } ( x ) = | x |\) and \(\mathrm { g } ( x ) = x + 1\), sketch the graphs of the composite functions \(y = \mathrm { fg } ( x )\) and \(y = \operatorname { gf } ( x )\), indicating clearly which is which.

2 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\) \begin{figure}[h] \end{figure}

1. Show algebraically that \(\mathrm { f } ( x )\) is an odd function. Interpret this result geometrically.
2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 2 } { \left( 2 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\). Hence find the exact gradient of the curve at the origin.
3. Find the exact area of the region bounded by the curve, the \(x\)-axis and the line \(x = 1\).
4. \(( A )\) Show that if \(y = \frac { x } { \sqrt { 2 + x ^ { 2 } } }\), then \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\).
   (B) Differentiate \(\frac { 1 } { y ^ { 2 } } = \frac { 2 } { x ^ { 2 } } + 1\) implicitly to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y ^ { 3 } } { x ^ { 3 } }\). Explain why this expression cannot be used to find the gradient of the curve at the origin.

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 \(C\) .

1.The function f is given by $$\mathrm { f } ( x ) = x ^ { 2 } - 2 x + 6 , \quad x \geqslant 0$$

1. Find the range of \(f\) . The function \(g\) is given by $$\mathrm { g } ( x ) = 3 + \sqrt { } ( x + 4 ) , \quad x \geqslant 2$$
2. Find \(\operatorname { gf } ( x )\) .
3. Find the domain and range of gf.

1.The function f is given by $$\mathrm { f } ( x ) = x ^ { 2 } - 4 x + 9 \quad x \in \mathbb { R } , x \geqslant 3$$

1. Find the range of f . The function g is given by $$\operatorname { g } ( x ) = \frac { 10 } { x + 1 } \quad x \in \mathbb { R } , x \geqslant 4$$
2. Find an expression for \(\operatorname { gf } ( x )\) .
3. Find the domain and range of gf.

1.The function f is given by $$\mathrm { f } ( x ) = \sqrt { x + 2 } \quad \text { for } \quad x \in \mathbb { R } , x \geqslant 0$$

1. Find \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of \(\mathrm { f } ^ { - 1 }\) The function g is given by $$\mathrm { g } ( x ) = x ^ { 2 } - 4 x + 5 \text { for } x \in \mathbb { R } , x \geqslant 0$$
2. Find the range of g .
3. Solve the equation \(\operatorname { fg } ( x ) = x\) .

9

1. The function f is defined for all real values of \(x\) by $$f ( x ) = e ^ { 2 x } - 3 e ^ { - 2 x } .$$
   1. Show that \(\mathrm { f } ^ { \prime } ( x ) > 0\) for all \(x\).
   2. Show that the set of values of \(x\) for which \(\mathrm { f } ^ { \prime \prime } ( x ) > 0\) is the same as the set of values of \(x\) for which \(\mathrm { f } ( x ) > 0\), and state what this set of values is.
   3. \includegraphics[max width=\textwidth, alt={}, center]{774bb427-5392-45d3-8e4e-47d08fb8a792-04_634_830_641_699} The function g is defined for all real values of \(x\) by $$\mathrm { g } ( x ) = \mathrm { e } ^ { 2 x } + k \mathrm { e } ^ { - 2 x } ,$$ where \(k\) is a constant greater than 1 . The graph of \(y = \mathrm { g } ( x )\) is shown above. Find the range of g , giving your answer in simplified form.

5 \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-3_844_837_242_621} It is given that f is a one-one function defined for all real values. The diagram shows the curve with equation \(y = \mathrm { f } ( x )\). The coordinates of certain points on the curve are shown in the following table.

1. State the value of \(\mathrm { ff } ( 6 )\) and the value of \(\mathrm { f } ^ { - 1 } ( 8 )\).
2. On the copy of the diagram, sketch the curve \(y = \mathrm { f } ^ { - 1 } ( x )\), indicating how the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) are related.
3. Use Simpson's rule with 6 strips to find an approximation to \(\int _ { 2 } ^ { 14 } \mathrm { f } ( x ) \mathrm { d } x\).

7 The functions \(\mathrm { f } , \mathrm { g }\) and h are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = | x | , \quad \mathrm { g } ( x ) = 3 x + 5 \quad \text { and } \quad \mathrm { h } ( x ) = \mathrm { gg } ( x ) .$$

1. Solve the equation \(\mathrm { g } ( x + 2 ) = \mathrm { f } ( - 12 )\).
2. Find \(\mathrm { h } ^ { - 1 } ( x )\).
3. Determine the values of \(x\) for which $$x + \mathrm { f } ( x ) = 0 .$$

7 The function f is defined for all real values of \(x\) by \(\mathrm { f } ( x ) = 2 x + 5\). The function g is defined for all real values of \(x\) and is such that \(\mathrm { g } ^ { - 1 } ( x ) = \sqrt [ 3 ] { x - a }\), where \(a\) is a constant. It is given that \(\mathrm { fg } ^ { - 1 } ( 12 ) = 9\). Find the value of \(a\) and hence solve the equation \(\operatorname { gf } ( x ) = 68\).

8 The functions f and g are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = | 2 x + a | + 3 a \quad \text { and } \quad \mathrm { g } ( x ) = 5 x - 4 a$$ where \(a\) is a positive constant.

1. State the range of f and the range of g .
2. State why f has no inverse, and find an expression for \(\mathrm { g } ^ { - 1 } ( x )\).
3. Solve for \(x\) the equation \(\operatorname { gf } ( x ) = 31 a\).

1. Show that \(\sin 2 \theta ( \tan \theta + \cot \theta ) \equiv 2\).
2. Hence
   1. find the exact value of \(\tan \frac { 1 } { 12 } \pi + \tan \frac { 1 } { 8 } \pi + \cot \frac { 1 } { 12 } \pi + \cot \frac { 1 } { 8 } \pi\),
   2. solve the equation \(\sin 4 \theta ( \tan \theta + \cot \theta ) = 1\) for \(0 < \theta < \frac { 1 } { 2 } \pi\),
   3. express \(( 1 - \cos 2 \theta ) ^ { 2 } \left( \tan \frac { 1 } { 2 } \theta + \cot \frac { 1 } { 2 } \theta \right) ^ { 3 }\) in terms of \(\sin \theta\).

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.

6 Write down the conditions for \(\mathrm { f } ( x )\) to be an odd function and for \(\mathrm { g } ( x )\) to be an even function.
Hence prove that, if \(\mathrm { f } ( x )\) is odd and \(\mathrm { g } ( x )\) is even, then the composite function \(\mathrm { gf } ( x )\) is even.