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

1 Functions f and g are defined for all real values of \(x\) by $$\mathrm { f } ( x ) = x ^ { 3 } + 4 \quad \text { and } \quad \mathrm { g } ( x ) = 2 x - 5$$ Evaluate

1. \(f g ( 1 )\),
2. \(\mathrm { f } ^ { - 1 } ( 12 )\).

1 The function f is defined for all real values of \(x\) by $$f ( x ) = 10 - ( x + 3 ) ^ { 2 } .$$

1. State the range of f .
2. Find the value of \(\mathrm { ff } ( - 1 )\).

6 \includegraphics[max width=\textwidth, alt={}, center]{ebfdf170-99c6-4785-b9d7-201c3425b4c9-3_563_583_267_781} The diagram shows the graph of \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = 2 - x ^ { 2 } , \quad x \leqslant 0 .$$

1. Evaluate \(\mathrm { ff } ( - 3 )\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. Sketch the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\). Indicate the coordinates of the points where the graph meets the axes.

3 The function \(f\) is defined for all non-negative values of \(x\) by $$f ( x ) = 3 + \sqrt { x }$$

1. Evaluate ff(169).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( \mathrm { x } )\) in terms of x .
3. On a single diagram sketch the graphs of \(y = f ( x )\) and \(y = f ^ { - 1 } ( x )\), indicating how the two graphs are related.

2 \includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-2_529_855_397_646} The diagram shows the graph of \(y = \mathrm { f } ( x )\). It is given that \(\mathrm { f } ( - 3 ) = 0\) and \(\mathrm { f } ( 0 ) = 2\). Sketch, on separate diagrams, the following graphs, indicating in each case the coordinates of the points where the graph crosses the axes:

1. \(y = \mathrm { f } ^ { - 1 } ( x )\),
2. \(y = - 2 \mathrm { f } ( x )\).

9 \includegraphics[max width=\textwidth, alt={}, center]{5c501214-b41c-43a8-b9c6-986758e83e7d-4_534_935_264_605} The function f is defined for the domain \(x \geqslant 0\) by $$f ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$ The diagram shows the curve with equation \(y = \mathrm { f } ( x )\).

1. Find the range of f .
2. The function g is defined for the domain \(x \geqslant k\) by $$\mathrm { g } ( x ) = \frac { 15 x } { x ^ { 2 } + 5 }$$ Given that g is a one-one function, state the least possible value of \(k\).
3. Show that there is no point on the curve \(y = \mathrm { g } ( x )\) at which the gradient is - 1 .

6 The function f is defined by $$\mathrm { f } : x \mapsto 1 + \sqrt { } x \quad \text { for } x \geqslant 0$$

1. State the domain and range of the inverse function \(\mathrm { f } ^ { - 1 }\).
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. By considering the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), show that the solution to the equation $$\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )$$ is \(x = \frac { 1 } { 2 } ( 3 + \sqrt { } 5 )\).

6 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 \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x - 1 } { 2 } \right)\) 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]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-4_506_561_705_751} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .

8 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 \(\mathrm { g } ( x ) = \ln ( 1 + \sqrt { x } )\) for \(x \geqslant 0\).
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 \(\int \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x\), 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 \(x = 1\).

6 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.

8 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.

2 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.

6 The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = 1 + 2 \sin 3 x , \quad - \frac { \pi } { 6 } \leqslant x \leqslant \frac { \pi } { 6 }$$ You are given that this function has an inverse, \(\mathrm { f } ^ { - 1 } ( x )\).
Find \(\mathrm { f } ^ { - 1 } ( x )\) and its domain.

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

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

3 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 \(\ln \frac { 32 } { 27 }\). [Hint: consider its reflection in \(y = x\).]

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.

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).

1 Fig. 4 shows the curve \(y = \mathrm { f } ( x )\), where $$f ( x ) = a + \cos b x , 0 \leqslant x \leqslant 2 \pi ,$$ and \(a\) and \(b\) are positive constants. The curve has stationary points at \(( 0,3 )\) and \(( 2 \pi , 1 )\). \begin{figure}[h] \end{figure}

1. Find \(a\) and \(b\).
2. Find \(\mathrm { f } ^ { - 1 } ( x )\), and state its domain and range.

2 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.

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 \(( 1,0 )\).

4 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = 2 \arcsin x , - 1 \nless \leqslant 1\).
Fig. 6 also shows the curve \(y = \mathrm { g } ( x )\), where \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
P is the point on the curve \(y = \mathrm { f } ( x )\) with \(x\)-coordinate \(\frac { 1 } { 2 }\).

1. Find the \(y\)-coordinate of P , giving your answer in terms of \(\pi\). The point Q is the reflection of P in \(y = x\).
2. Find \(\mathrm { g } ( x )\) and its derivative \(\mathrm { g } ^ { \prime } ( x )\). Hence determine the exact gradient of the curve \(y = \mathrm { g } ( x )\) at the point Q . Write down the exact gradient of \(y = \mathrm { f } ( x )\) at the point P .

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 \(x = \pi\).