Sketch function and inverse graphs

11 \includegraphics[max width=\textwidth, alt={}, center]{b24ed4c7-ab07-45f4-adf2-027734c36b62-4_862_892_932_628} The diagram shows the graph of \(y = \mathrm { f } ( x )\), where \(\mathrm { f } : x \mapsto \frac { 6 } { 2 x + 3 }\) for \(x \geqslant 0\).

1. Find an expression, in terms of \(x\), for \(\mathrm { f } ^ { \prime } ( x )\) and explain how your answer shows that f is a decreasing function.
2. Find an expression, in terms of \(x\), for \(\mathrm { f } ^ { - 1 } ( x )\) and find the domain of \(\mathrm { f } ^ { - 1 }\).
3. Copy the diagram and, on your copy, sketch the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\), making clear the relationship between the graphs. The function g is defined by \(\mathrm { g } : x \mapsto \frac { 1 } { 2 } x\) for \(x \geqslant 0\).
4. Solve the equation \(\operatorname { fg } ( x ) = \frac { 3 } { 2 }\).

11 Functions \(f\) and \(g\) are defined by $$\begin{array} { l l } \mathrm { f } : x \mapsto 2 x ^ { 2 } - 8 x + 10 & \text { for } 0 \leqslant x \leqslant 2 \\ \mathrm {~g} : x \mapsto x & \text { for } 0 \leqslant x \leqslant 10 \end{array}$$

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. State the domain of \(\mathrm { f } ^ { - 1 }\).
4. Sketch on the same diagram the graphs of \(y = \mathrm { f } ( x ) , y = \mathrm { g } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), making clear the relationship between the graphs.
5. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).

2 A function f is defined by \(\mathrm { f } : x \mapsto 4 - 5 x\) for \(x \in \mathbb { R }\).

1. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and find the point of intersection of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\).
2. Sketch, on the same diagram, the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\), making clear the relationship between the graphs.

1. The function f is defined by

$$\mathrm { f } ( x ) = 2 x ^ { 2 } - 5 \quad x \geqslant 0 \quad x \in \mathbb { R }$$

1. State the range of f On the following page there is a diagram, labelled Diagram 1, which shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\).
2. On Diagram 1, sketch the curve with equation \(y = \mathrm { f } ^ { - 1 } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) meets the curve with equation \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(P\) Using algebra and showing your working,
3. find the exact \(x\) coordinate of \(P\)
   \includegraphics[max width=\textwidth, alt={}]{bef290fb-fbac-4c9c-981e-5e323ac7182e-09_607_610_248_731}
   \section*{Diagram 1}

1. $$g ( x ) = \frac { 6 x + 12 } { x ^ { 2 } + 3 x + 2 } - 2 , \quad x \geqslant 0$$

1. Show that \(\mathrm { g } ( x ) = \frac { 4 - 2 x } { x + 1 } , x \geqslant 0\)
2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0f6fd881-4d4b-4f80-96cc-6da41cc33c60-02_494_922_628_511} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { g } ( x ) , x \geqslant 0\) The curve meets the \(y\)-axis at \(( 0,4 )\) and crosses the \(x\)-axis at \(( 2,0 )\). On separate diagrams sketch the graph with equation
   1. \(y = 2 \mathrm {~g} ( 2 x )\),
   2. \(y = \mathrm { g } ^ { - 1 } ( x )\). Show on each sketch the coordinates of each point at which the graph meets or crosses the axes.

9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\). \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 \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
4. Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(\mathrm { g } ( x )\). Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).

\includegraphics{figure_8} The diagram shows the graph of \(y = f(x)\) where the function \(f\) is defined by $$f(x) = 3 + 2\sin \frac{1}{4}x \text{ for } 0 \leqslant x \leqslant 2\pi.$$

1. On the diagram above, sketch the graph of \(y = f^{-1}(x)\). [2]
2. Find an expression for \(f^{-1}(x)\). [2]
3. \includegraphics{figure_8c} The diagram above shows part of the graph of the function \(g(x) = 3 + 2\sin \frac{1}{4}x\) for \(-2\pi \leqslant x \leqslant 2\pi\). Complete the sketch of the graph of \(g(x)\) on the diagram above and hence explain whether the function \(g\) has an inverse. [2]
4. Describe fully a sequence of three transformations which can be combined to transform the graph of \(y = \sin x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\) to the graph of \(y = f(x)\), making clear the order in which the transformations are applied. [6]

Write down the inverse function of \(y = e^x\). On the same set of axes, sketch the graphs of \(y = e^x\) and its inverse function, clearly labelling the coordinates of the points where the graphs cross the \(x\) and \(y\) axes. [3]

% Figure shows a curve with maximum at point A, passing through origin O, with horizontal asymptote \includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = f(x)\) where $$f(x) = \frac{x^2 + 16}{3x} \quad x \neq 0$$ The curve has a maximum at the point \(A\) with coordinates \((a, b)\).

1. Find the value of \(a\) and the value of \(b\). [4] The function g is defined as $$g : x \mapsto \frac{x^2 + 16}{3x} \quad a \leq x < 0$$ where \(a\) is the value found in part (a).
2. Write down the range of g. [1]
3. On the same axes sketch \(y = g(x)\) and \(y = g^{-1}(x)\). [3]
4. Find an expression for \(g^{-1}(x)\) and state the domain of \(g^{-1}\) [5]
5. Solve the equation \(g(x) = g^{-1}(x)\). [3]