Unit 9 Review Exercises

Functions

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 Let \(f(x) = 2x^2 - 3x\). Find

(a) \(f(-2)\)

(b) \(f(x - 1)\), simplified

(c) the values of \(x\) for which \(f(x) = 2\).

Solution

(a) \(8 + 6 = 14\).

(b) \(2(x - 1)^2 - 3(x - 1) = 2x^2 - 4x + 2 - 3x + 3 = 2x^2 - 7x + 5\).

(c) \(2x^2 - 3x - 2 = 0 \Rightarrow (2x + 1)(x - 2) = 0 \Rightarrow x = -\dfrac{1}{2}\) or \(x = 2\). See 9.1.

2 Paper 1 Find the largest possible domain of each function:

(a) \(f(x) = \dfrac{4}{2x + 6}\)

(b) \(g(x) = \sqrt{10 - 2x}\)

(c) \(h(x) = \ln(x^2 - 4)\).

Solution

(a) \(x \neq -3\).

(b) \(10 - 2x \geq 0 \Rightarrow x \leq 5\).

(c) \(x^2 > 4 \Rightarrow x < -2\) or \(x > 2\). See 9.2.

3 Paper 1 Find the range of \(f(x) = x^2 + 6x + 4\) for

(a) \(x \in \mathbb{R}\),

(b) \(-1 \leq x \leq 2\).

Solution

\(f(x) = (x + 3)^2 - 5\).

(a) \(f(x) \geq -5\).

(b) The vertex \(x = -3\) is outside the domain, and \(f\) is increasing for \(x > -3\). \(f(-1) = -1\), \(f(2) = 20\), so \(-1 \leq f(x) \leq 20\). See 9.2.

4 Paper 1 Let \(f(x) = \dfrac{3x - 2}{x + 4}\), \(x \neq -4\).

(a) Find \(f^{-1}(x)\) and state its domain.

(b) Hence, or otherwise, solve \(f(x) = 1\).

Solution

(a) \(y(x + 4) = 3x - 2 \Rightarrow xy - 3x = -4y - 2 \Rightarrow x = \dfrac{-4y - 2}{y - 3} = \dfrac{4y + 2}{3 - y}\).

\(f^{-1}(x) = \dfrac{4x + 2}{3 - x}\), \(\;x \neq 3\).

(b) \(x = f^{-1}(1) = \dfrac{6}{2} = 3\). (Check: \(f(3) = \dfrac{7}{7} = 1\) ✓.) See 9.3.

5 Paper 1 Let \(f(x) = x^2 - 1\) and \(g(x) = 3 - 2x\), both for \(x \in \mathbb{R}\).

(a) Find \((f \circ g)(x)\) and \((g \circ f)(x)\).

(b) Solve \((f \circ g)(x) = (g \circ f)(x)\).

Solution

(a) \((f \circ g)(x) = (3 - 2x)^2 - 1 = 4x^2 - 12x + 8\). \(\quad (g \circ f)(x) = 3 - 2(x^2 - 1) = 5 - 2x^2\).

(b) \(4x^2 - 12x + 8 = 5 - 2x^2 \Rightarrow 6x^2 - 12x + 3 = 0 \Rightarrow 2x^2 - 4x + 1 = 0\).

\[ x = \frac{4 \pm \sqrt{8}}{4} = 1 \pm \frac{\sqrt{2}}{2} . \]

See 9.4.

6 Paper 1 Let \(f(x) = 4x + 1\) and \(g(x) = \dfrac{x - 1}{4}\). Show that \(g\) is the inverse of \(f\) by finding \((f \circ g)(x)\) and \((g \circ f)(x)\).

Solution

\((f \circ g)(x) = 4 \cdot \dfrac{x - 1}{4} + 1 = x\) and \((g \circ f)(x) = \dfrac{4x + 1 - 1}{4} = x\). Both composites are the identity, so \(g = f^{-1}\). See 9.4.

7 Paper 1 AHL The function \(f(x) = 5 - 4x - x^2\) has domain \(x \leq a\).

(a) Find the greatest value of \(a\) for which \(f^{-1}\) exists.

(b) For this value of \(a\), find \(f^{-1}(x)\) and its domain.

Solution

(a) \(f(x) = 9 - (x + 2)^2\). The vertex is at \(x = -2\), so the greatest value is \(a = -2\).

(b) \((x + 2)^2 = 9 - y\), and \(x \leq -2\) means \(x + 2 \leq 0\): \(\;x = -2 - \sqrt{9 - y}\).

\(f^{-1}(x) = -2 - \sqrt{9 - x}\), \(\;x \leq 9\) (the range of \(f\)). See 9.5.

8 Paper 2 Let \(f(x) = \mathrm{e}^{x} - 2\), \(x \in \mathbb{R}\).

(a) Find \(f^{-1}(x)\) and state its domain.

(b) Sketch \(y = f(x)\) and \(y = f^{-1}(x)\) on one diagram, showing their asymptotes.

(c) Use your GDC to find the coordinates of the points where the two graphs meet.

Solution

(a) \(\mathrm{e}^{x} = y + 2 \Rightarrow f^{-1}(x) = \ln(x + 2)\), \(\;x > -2\).

(b) \(y = f(x)\): horizontal asymptote \(y = -2\), through \((0, -1)\) and \((\ln 2, 0)\). \(y = f^{-1}(x)\): vertical asymptote \(x = -2\), through \((-1, 0)\) and \((0, \ln 2)\). The two are reflections in \(y = x\).

(c) \(f\) is increasing, so the graphs meet on \(y = x\): solve \(\mathrm{e}^{x} - 2 = x\). The GDC gives \(x = -1.84\) and \(x = 1.15\) (3 s.f.). The points are \((-1.84, -1.84)\) and \((1.15, 1.15)\). See 9.3.

9 Paper 1 AHL

(a) Show that \(f(x) = \dfrac{x^3}{x^4 + 1}\) is odd.

(b) The function \(g(x) = (x - 2)\left(\dfrac{1}{x} + b\right)\), \(x \neq 0\), is odd. Find \(b\).

Solution

(a) \(f(-x) = \dfrac{-x^3}{x^4 + 1} = -f(x)\), so \(f\) is odd.

(b) \(g(x) = 1 + bx - \dfrac{2}{x} - 2b = (1 - 2b) + bx - \dfrac{2}{x}\). The constant must be zero: \(1 - 2b = 0\), \(b = \dfrac{1}{2}\). See 9.5.

10 Paper 1 AHL Let \(f(x) = \dfrac{ax + 3}{x - 4}\), \(x \neq 4\).

(a) Find the value of \(a\) for which \(f\) is self-inverse.

(b) For this value of \(a\), find \((f \circ f)(7)\) without any working, and explain your answer.

Solution

(a) \(y(x - 4) = ax + 3 \Rightarrow x(y - a) = 4y + 3 \Rightarrow f^{-1}(x) = \dfrac{4x + 3}{x - a}\). For \(f^{-1} = f\) we need \(a = 4\).

(b) \(f\) is self-inverse, so \((f \circ f)(x) = x\) for every \(x\) in the domain. Hence \((f \circ f)(7) = 7\). (Check: \(f(7) = \dfrac{31}{3}\) and \(f\!\left(\dfrac{31}{3}\right) = \dfrac{\frac{124}{3} + 3}{\frac{31}{3} - 4} = \dfrac{133}{19} = 7\) ✓.) See 9.5.

11 Paper 1 Let \(f(x) = -x^2 + 2x + 8\). Find the zeros of \(f\), the \(y\)-intercept and the coordinates of the vertex. Sketch the graph of \(y = f(x)\), labelling these features.

Solution

\(-x^2 + 2x + 8 = -(x^2 - 2x - 8) = -(x - 4)(x + 2)\), so the zeros are \(x = -2\) and \(x = 4\). The \(y\)-intercept is \((0, 8)\).

Axis of symmetry \(x = -\dfrac{2}{2 \times (-1)} = 1\); \(\;f(1) = -1 + 2 + 8 = 9\). The vertex is the maximum point \((1, 9)\).

The sketch is a parabola opening downwards through \((-2, 0)\), \((0, 8)\), \((1, 9)\) and \((4, 0)\). See 9.6 and 8.2.

12 Paper 1 Solve \(4^x - 6 \times 2^x + 8 = 0\).

Solution

Let \(u = 2^x\), so \(4^x = u^2\): \(\;u^2 - 6u + 8 = (u - 2)(u - 4) = 0\).

\(2^x = 2 \Rightarrow x = 1\); \(\;2^x = 4 \Rightarrow x = 2\). See 9.7.

13 Paper 2 Find the coordinates of the points of intersection of \(y = \mathrm{e}^{0.5x}\) and \(y = 4 - x^2\).

Solution

Using Analyze Graph → Intersection twice: \((-1.90,\, 0.387)\) and \((1.41,\, 2.02)\). See 9.7.