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

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 \(\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }\).

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point \(( 0,1 )\). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

2 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 \(y = \mathrm { f } ( x )\) ?

3 The function \(f ( x ) = \ln \left( t + x ^ { 2 } \right)\) 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.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. 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 \(g ( x )\).
4. Sketch the curves \(y = f ( x )\) and \(y = g ( x )\) on the same axes. State the domain of the function \(g ( x )\),
   Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( \mathrm { x } )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

1 Given that \(\mathrm { f } ( x ) = \frac { x + 1 } { x - 1 }\), show that \(\mathrm { ff } ( x ) = x\).
Hence write down the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). What can you deduce about the symmetry of the curve \(y = \mathrm { f } ( x ) ?\)

2 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for all real numbers \(x\) by $$\mathrm { f } ( x ) = x ^ { 2 } , \quad \mathrm {~g} ( x ) = x - 2 .$$

1. Find the composite functions \(\mathrm { fg } ( x )\) and \(\mathrm { gf } ( x )\).
2. Sketch the curves \(y = \mathrm { f } ( x ) , y = \mathrm { fg } ( x )\) and \(y = \mathrm { gf } ( x )\), indicating clearly which is which.

3 Given that \(\mathrm { f } ( x ) = 1 - x\) and \(\mathrm { g } ( x ) = | x |\), write down the composite function \(\mathrm { gf } ( x )\). On separate diagrams, sketch the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { gf } ( x )\).

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 .

6 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for the domain \(x > 0\) as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function \(\mathrm { fg } ( x )\) in terms of \(\ln x\).
State the transformation which maps the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { fg } ( x )\).

7 Fig. 9 shows the line \(y = x\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)\). The line and the curve intersect at the origin and at the point \(\mathrm { P } ( a , a )\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { e } ^ { a } = 1 + 2 a\).
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac { 1 } { 2 } a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\).
3. Show that the inverse function of \(\mathrm { f } ( x )\) is \(\mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = \ln ( 1 + 2 x )\). Add a sketch of \(y = \mathrm { g } ( x )\) to the copy of Fig. 9.
4. Find the derivatives of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }\). Give a geometrical interpretation of this result.

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 \(( 1,0 )\).

1 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P . [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R.
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\).
   [0pt] [Hint: consider its reflection in \(y = x\).]

2 Fig. 8 shows the line \(y = x\) and parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where $$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$ The curves intersect the axes at the points A and B, as shown. The curves and the line \(y = x\) meet at the point C . \begin{figure}[h] \end{figure}

1. Find the exact coordinates of A and B . Verify that the coordinates of C are \(( 1,1 )\).
2. Prove algebraically that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).
3. Evaluate \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\), giving your answer in terms of e .
4. Use integration by parts to find \(\int \ln x \mathrm {~d} x\). Hence show that \(\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }\).
5. Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .

3 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 1 + \sin 2 x\) for \(- \frac { 1 } { 4 } \pi \leqslant x \leqslant \frac { 1 } { 4 } \pi\). \begin{figure}[h] \end{figure}

1. State a sequence of two transformations that would map part of the curve \(y = \sin x\) onto the curve \(y = \mathrm { f } ( x )\).
2. Find the area of the region enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis and the line \(x = \frac { 1 } { 4 } \pi\).
3. Find the gradient of the curve \(y = \mathrm { f } ( x )\) at the point ( 0,1 ). Hence write down the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).
4. State the domain of \(\mathrm { f } ^ { - 1 } ( x )\). Add a sketch of \(y = \mathrm { f } ^ { - 1 } ( x )\) to a copy of Fig. 8.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

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.

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.

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.

8 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.

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

6 \includegraphics[max width=\textwidth, alt={}, center]{c940af95-e291-402a-856c-9090d13163d5-3_627_689_264_726} The function f is defined for all real values of \(x\) by $$f ( x ) = \sqrt [ 3 ] { \frac { 1 } { 2 } x + 2 }$$ The graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) meet at the point \(P\), and the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\) meets the \(x\)-axis at \(Q\) (see diagram).

1. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and determine the \(x\)-coordinate of the point \(Q\).
2. State how the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) are related geometrically, and hence show that the \(x\)-coordinate of the point \(P\) is the root of the equation $$x = \sqrt [ 3 ] { \frac { 1 } { 2 } x + 2 }$$
3. Use an iterative process, based on the equation \(x = \sqrt [ 3 ] { \frac { 1 } { 2 } x + 2 }\), to find the \(x\)-coordinate of \(P\), giving your answer correct to 2 decimal places.

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 \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 \(\int _ { 2 } ^ { 14 } \mathrm { f } ( x ) \mathrm { d } x\).