Find inverse function

The function f is defined by \(\mathrm{f}(x) = \frac{2x + 1}{2x - 1}\) for \(x < \frac{1}{2}\).

1. 1. State the value of f\((-1)\). [1]
   2. \includegraphics{figure_5} The diagram shows the graph of \(y = \mathrm{f}(x)\). Sketch the graph of \(y = \mathrm{f}^{-1}(x)\) on this diagram. Show any relevant mirror line. [2]
   3. Find an expression for \(\mathrm{f}^{-1}(x)\) and state the domain of the function \(\mathrm{f}^{-1}\). [4]

The function g is defined by \(\mathrm{g}(x) = 3x + 2\) for \(x \in \mathbb{R}\).

1. Solve the equation \(\mathrm{f}(x) = \mathrm{gf}\left(\frac{1}{4}\right)\). [3]

Functions \(f\) and \(g\) are defined by $$f : x \mapsto 2x + 5 \quad \text{for } x \in \mathbb{R},$$ $$g : x \mapsto \frac{8}{x - 3} \quad \text{for } x \in \mathbb{R}, x \neq 3.$$

1. Obtain expressions, in terms of \(x\), for \(f^{-1}(x)\) and \(g^{-1}(x)\), stating the value of \(x\) for which \(g^{-1}(x)\) is not defined. [4]
2. Sketch the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) on the same diagram, making clear the relationship between the two graphs. [3]
3. Given that the equation \(fg(x) = 5 - kx\), where \(k\) is a constant, has no solutions, find the set of possible values of \(k\). [5]

$$f(x) = \frac{2}{x-1} - \frac{6}{(x-1)(2x+1)}, \quad x > 1.$$

1. Prove that \(f(x) = \frac{4}{2x+1}\). [4]
2. Find the range of \(f\). [2]
3. Find \(f^{-1}(x)\). [3]
4. Find the range of \(f^{-1}(x)\). [1]

The function \(f\) is given by \(f: x \mapsto 2 + \frac{3}{x + 2}, x \in \mathbb{R}, x \neq -2\).

1. Express \(2 + \frac{3}{x + 2}\) as a single fraction. [1]
2. Find an expression for \(f^{-1}(x)\). [3]
3. Write down the domain of \(f^{-1}\). [1]

$$f(x) = \frac{2}{x - 1} - \frac{6}{(x - 1)(2x + 1)}, \quad x > 1$$

1. Prove that f(x) = \(\frac{4}{2x + 1}\). [4]
2. Find the range of f. [2]
3. Find \(f^{-1}(x)\). [3]
4. Find the range of \(f^{-1}(x)\). [1]

The function f is given by $$f: x \mapsto \ln(3x - 6), \quad x \in \mathbb{R}, \quad x > 2$$

1. Find \(f^{-1}(x)\). [3]
2. Write down the domain of \(f^{-1}\) and the range of \(f^{-1}\). [2]
3. Find, to 3 significant figures, the value of \(x\) for which f(x) = 3. [2]

The function g is given by $$g: x \mapsto \ln|3x - 6|, \quad x \in \mathbb{R}, \quad x \neq 2$$

1. Sketch the graph of \(y = g(x)\). [3]
2. Find the exact coordinates of all the points at which the graph of \(y = g(x)\) meets the coordinate axes. [3]

The function f is given by \(f: x \mapsto 2 + \frac{3}{x + 2}\), \(x \in \mathbb{R}\), \(x \neq -2\).

1. Express \(2 + \frac{3}{x + 2}\) as a single fraction. [1]
2. Find an expression for \(f^{-1}(x)\). [3]
3. Write down the domain of \(f^{-1}\). [1]

\includegraphics{figure_9} The function f is defined by \(f(x) = \sqrt{(mx + 7)} - 4\), where \(x \geq -\frac{7}{m}\) and \(m\) is a positive constant. The diagram shows the curve \(y = f(x)\).

1. A sequence of transformations maps the curve \(y = \sqrt{x}\) to the curve \(y = f(x)\). Give details of these transformations. [4]
2. Explain how you can tell that f is a one-one function and find an expression for \(f^{-1}(x)\). [4]
3. It is given that the curves \(y = f(x)\) and \(y = f^{-1}(x)\) do not meet. Explain how it can be deduced that neither curve meets the line \(y = x\), and hence determine the set of possible values of \(m\). [5]

\includegraphics{figure_6} The diagram shows the graph of \(y = f(x)\), where $$f(x) = 2 - x^2, \quad x \leq 0.$$

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

\includegraphics{figure_4} The function \(f\) is defined for all real values of \(x\) by $$f(x) = 2 - \sqrt{x + 1}.$$ The diagram shows the graph of \(y = f(x)\).

1. Evaluate \(f(-126)\). [2]
2. Find the set of values of \(x\) for which \(f(x) = |f(x)|\). [2]
3. Find an expression for \(f^{-1}(x)\). [3]
4. State how the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) are related geometrically. [1]

Fig. 7 shows the curve \(y = f(x)\), where \(f(x) = 1 + 2 \arctan x\), \(x \in \mathbb{R}\). The scales on the \(x\)- and \(y\)-axes are the same. \includegraphics{figure_7}

1. Find the range of f, giving your answer in terms of \(\pi\). [3]
2. Find \(f^{-1}(x)\), and add a sketch of the curve \(y = f^{-1}(x)\) to the copy of Fig. 7. [5]

Fig. 9 shows the curve \(y = f(x)\). The endpoints of the curve are P \((-\pi, 1)\) and Q \((\pi, 3)\), and \(f(x) = a + \sin bx\), where \(a\) and \(b\) are constants. \includegraphics{figure_9}

1. Using Fig. 9, show that \(a = 2\) and \(b = \frac{1}{2}\). [3]
2. Find the gradient of the curve \(y = f(x)\) at the point \((0, 2)\). Show that there is no point on the curve at which the gradient is greater than this. [5]
3. Find \(f^{-1}(x)\), and state its domain and range. Write down the gradient of \(y = f^{-1}(x)\) at the point \((2, 0)\). [6]
4. Find the area enclosed by the curve \(y = f(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \pi\). [4]

Fig. 4 shows the curve \(y = f(x)\), where $$f(x) = a + \cos bx, \quad 0 \leq x \leq 2\pi,$$ and \(a\) and \(b\) are positive constants. The curve has stationary points at \((0, 3)\) and \((2\pi, 1)\). \includegraphics{figure_4}

1. Find \(a\) and \(b\). [2]
2. Find \(f^{-1}(x)\), and state its domain and range. [5]

The functions f(x) and g(x) are defined by $$f(x) = x^2, \quad g(x) = 2x - 1,$$ for all real values of \(x\).

1. State the ranges of f(x) and g(x). Explain why f(x) has no inverse. [3]
2. Find an expression for the inverse function g\(^{-1}\)(x) in terms of \(x\). Sketch the graphs of \(y = g(x)\) and \(y = g^{-1}(x)\) on the same axes. [4]
3. Find expressions for gf(x) and fg(x). [2]
4. Solve the equation gf(x) = fg(x). Sketch the graphs of \(y = gf(x)\) and \(y = fg(x)\) on the same axes to illustrate your answer. [4]
5. Show that the equation f(x + a) = g\(^{-1}\)(x) has no solution if \(a > \frac{1}{4}\). [5]

The function f is defined by $$f : x \to 3e^{x-1}, \quad x \in \mathbb{R}.$$

1. State the range of f. [1]
2. Find an expression for \(f^{-1}(x)\) and state its domain. [4]

The function g is defined by $$g : x \to 5x - 2, \quad x \in \mathbb{R}.$$ Find, in terms of e,

1. the value of gf(ln 2), [3]
2. the solution of the equation $$f^{-1}g(x) = 4.$$ [4]

The function f is defined by $$\text{f}(x) \equiv 2 + \ln (3x - 2), \quad x \in \mathbb{R}, \quad x > \frac{2}{3}.$$

1. Find the exact value of \(\text{f}(1)\). [2]
2. Find an equation for the tangent to the curve \(y = \text{f}(x)\) at the point where \(x = 1\). [4]
3. Find an expression for \(\text{f}^{-1}(x)\). [2]

The function f is defined by $$\text{f}(x) \equiv 3 - x^2, \quad x \in \mathbb{R}, \quad x \geq 0.$$

1. State the range of f. [1]
2. Sketch the graphs of \(y = \text{f}(x)\) and \(y = \text{f}^{-1}(x)\) on the same diagram. [3]
3. Find an expression for f\(^{-1}(x)\) and state its domain. [3]

The function g is defined by $$\text{g}(x) \equiv \frac{8}{3-x}, \quad x \in \mathbb{R}, \quad x \neq 3.$$

1. Evaluate fg\((-3)\). [2]
2. Solve the equation $$\text{f}^{-1}(x) = \text{g}(x).$$ [3]

Fig. 9 shows the curve \(y = f(x)\), where $$f(x) = (e^x - 2)^2 - 1, x \in \mathbb{R}.$$ The curve crosses the x-axis at O and P, and has a turning point at Q. \includegraphics{figure_9}

1. Find the exact x-coordinate of P. [2]
2. Show that the x-coordinate of Q is \(\ln 2\) and find its y-coordinate. [4]
3. Find the exact area of the region enclosed by the curve and the x-axis. [5]

The domain of f(x) is now restricted to \(x \geqslant \ln 2\).

1. Find the inverse function \(f^{-1}(x)\). Write down its domain and range, and sketch its graph on the copy of Fig. 9. [7]

Fig. 7 shows the curve \(y = f(x)\), where \(f(x) = 1 + 2\arctan x, x \in \mathbb{R}\). The scales on the x- and y-axes are the same. \includegraphics{figure_7}

1. Find the range of f, giving your answer in terms of \(\pi\). [3]
2. Find \(f^{-1}(x)\), and add a sketch of the curve \(y = f^{-1}(x)\) to the copy of Fig. 7. [5]

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

$$f(x) = \frac{ax + b}{x + 2}; \quad x \in \mathbb{R}, x \neq -2,$$ where \(a\) and \(b\) are constants and \(b > 0\).

1. Find \(f^{-1}(x)\). [2]
2. Hence, or otherwise, find the value of \(a\) so that \(f(x) = x\). [2]

The curve \(C\) has equation \(y = f(x)\) and \(f(x)\) satisfies \(f(x) = x\).

1. On separate axes sketch
   1. \(y = f(x)\), [3]
   2. \(y = f(x - 2) + 2\). [3]

On each sketch you should indicate the equations of any asymptotes and the coordinates, in terms of \(b\), of any intersections with the axes. The normal to \(C\) at the point \(P\) has equation \(y = 4x - 39\). The normal to \(C\) at the point \(Q\) has equation \(y = 4x + k\), where \(k\) is a constant.

1. By considering the images of the normals to \(C\) on the curve with equation \(y = f(x - 2) + 2\), or otherwise, find the value of \(k\). [5]

1. Give full details of the single transformation that transforms the graph of \(y = x^3\) to the graph of \(y = x^3 - 8\). [2]

The function f is defined by \(\mathrm{f}(x) = x^3 - 8\).

1. Find an expression for \(\mathrm{f}^{-1}(x)\). [2]
2. State how the graphs of \(y = \mathrm{f}(x)\) and \(y = \mathrm{f}^{-1}(x)\) are related geometrically. [1]

The function f is defined by $$f(x) = \frac{1}{2}(x^2 + 1), \quad x \geq 0$$

1. Find the range of f. [1 mark]
2. 1. Find \(f^{-1}(x)\) [3 marks]
   2. State the range of \(f^{-1}(x)\) [1 mark]
3. State the transformation which maps the graph of \(y = f(x)\) onto the graph of \(y = f^{-1}(x)\) [1 mark]
4. Find the coordinates of the point of intersection of the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) [2 marks]