Functions and Transformations: practice solutions, Grade 11

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Practice solutions Grade 11 : Functions and Transformations — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

Sommaire

1 Evaluating a function ★★★

Substitute each input in parentheses.

$ f(-2) = 3(4) - 2(-2) + 5 = 12 + 4 + 5 = 21 $.

$ f(0) = 0 - 0 + 5 = 5 $.

$ f(3) = 3(9) - 6 + 5 = 27 - 6 + 5 = 26 $.

Answer: $ f(-2) = 21 $, $ f(0) = 5 $, $ f(3) = 26 $.

2 Is it a function? ★★★

a) In $ A $, every first coordinate appears once, so each input has one output: $ A $ is a function. (Two inputs may share an output, as $ -1 $ and $ 0 $ do.) In $ B $, the input 1 is paired with both 3 and $ -3 $, so $ B $ is not a function.

b) Domain of $ A $: $ \{-3, -1, 0, 2, 4\} $. Range: $ \{-5, 1, 2, 4\} $ (the value 2 is listed once).

3 Finding domains ★★★

a) Need $ x - 4 \ge 0 $, so $ x \ge 4 $.

b) The denominator cannot be zero: $ x + 3 \ne 0 $, so $ x \ne -3 $.

c) Need $ 3x + 12 \ge 0 $, so $ 3x \ge -12 $ and $ x \ge -4 $.

4 Name the shift ★★★

a) Parent $ x^2 $: shifted 4 units left and 7 units down. Vertex $ (-4, -7) $.

b) Parent $ |x| $: reflected over the x-axis, then shifted 2 units up. Vertex $ (0, 2) $, opening downward.

c) Parent $ \sqrt{x} $: shifted 5 units right. The starting point is $ (5, 0) $.

5 Sum, difference, product ★★★

$ (f+g)(x) = 2x + 3 + x - 6 = 3x - 3 $.

$ (f-g)(x) = 2x + 3 - x + 6 = x + 9 $.

$ (fg)(x) = (2x+3)(x-6) = 2x^2 - 12x + 3x - 18 = 2x^2 - 9x - 18 $.

$ (f+g)(4) = 3(4) - 3 = 9 $. Check: $ f(4) + g(4) = 11 + (-2) = 9 $.

6 First composition ★★★

a) $ g(3) = 6 $, so $ f(g(3)) = f(6) = 11 $. $ f(3) = 8 $, so $ g(f(3)) = g(8) = 16 $.

b) $ f(g(x)) = 2x + 5 $ and $ g(f(x)) = 2(x+5) = 2x + 10 $. They are not equal (they differ by 5), so the order matters.

7 A first inverse ★★★

Write $ y = 4x - 7 $ and swap: $ x = 4y - 7 $. Solve: $ 4y = x + 7 $, so $ y = \dfrac{x+7}{4} $. Thus $ f^{-1}(x) = \dfrac{x + 7}{4} $.

Check: $ f^{-1}(5) = \dfrac{12}{4} = 3 $, and indeed $ f(3) = 4(3) - 7 = 5 $.

8 Piecewise evaluation ★★★

Find the piece for each input.

$ -3 \lt -1 $: $ f(-3) = -3 + 4 = 1 $.

$ x = -1 $ belongs to the middle piece: $ f(-1) = (-1)^2 = 1 $.

$ f(0) = 0^2 = 0 $.

$ x = 2 $ belongs to the middle piece (since $ 2 \le 2 $): $ f(2) = 4 $.

$ 5 \gt 2 $: $ f(5) = 6 - 5 = 1 $.

9 Build the equation ★★★

a) Reflection: $ -x^2 $. Shift right 3: replace $ x $ by $ x - 3 $. Shift up 5: add 5. So $ g(x) = -(x-3)^2 + 5 $.

b) The vertex is $ (3, 5) $. $ g(1) = -(1-3)^2 + 5 = -4 + 5 = 1 $.

10 Taxi fare ★★★

a) A distance cannot be negative: domain $ m \ge 0 $. The smallest fare is $ C(0) = 3 $, so the range is $ C \ge 3 $.

b) $ C(12) = 3 + 2.5(12) = 33 $: the fare is \$33.

c) Solve $ 3 + 2.5m = 48 $: $ 2.5m = 45 $, so $ m = 18 $ miles.

d) Swap and solve: $ m = \dfrac{C - 3}{2.5} $, so $ C^{-1}(x) = \dfrac{x - 3}{2.5} $. It converts a fare in dollars into the distance in miles that was driven (check: $ C^{-1}(48) = 18 $).

11 Composition with a quadratic ★★★

$ f(g(x)) = (2x+1)^2 - 3 = 4x^2 + 4x + 1 - 3 = 4x^2 + 4x - 2 $.

$ g(f(x)) = 2(x^2 - 3) + 1 = 2x^2 - 5 $.

$ f(g(2)) = f(5) = 25 - 3 = 22 $; the formula gives $ 16 + 8 - 2 = 22 $.

$ g(f(2)) = g(1) = 3 $; the formula gives $ 8 - 5 = 3 $.

12 A quotient with a hole ★★★

a) $ \dfrac{x^2-9}{x-3} = \dfrac{(x-3)(x+3)}{x-3} = x + 3 $. The domain still excludes the value that makes $ g(x) = 0 $: $ x \ne 3 $. The graph is the line $ y = x + 3 $ with a hole at $ (3, 6) $.

b) $ \dfrac{g}{f} = \dfrac{1}{x+3} $ after simplifying, but $ f(x) = 0 $ when $ x = 3 $ or $ x = -3 $. The domain is $ x \ne 3 $ and $ x \ne -3 $.

13 Inverse of a rational function ★★★

Write $ y = \dfrac{3}{x-1} + 2 $ and swap: $ x = \dfrac{3}{y-1} + 2 $. Then $ x - 2 = \dfrac{3}{y-1} $, so $ y - 1 = \dfrac{3}{x-2} $ and $ y = \dfrac{3}{x-2} + 1 $. Thus $ f^{-1}(x) = \dfrac{3}{x-2} + 1 $.

Check: $ f(4) = \dfrac{3}{3} + 2 = 3 $ and $ f^{-1}(3) = \dfrac{3}{1} + 1 = 4 $.

14 Absolute value equation and inequality ★★★

a) $ 2x - 5 = 9 $ gives $ x = 7 $; $ 2x - 5 = -9 $ gives $ x = -2 $. Check: $ |2(-2) - 5| = |-9| = 9 $.

b) $ |x-3| \lt 4 $ means “the distance from $ x $ to 3 is less than 4”, so $ -4 \lt x - 3 \lt 4 $, that is $ -1 \lt x \lt 7 $. Interval notation: $ (-1, 7) $.

15 Domain and range from a transformation ★★★

a) The radicand must be nonnegative: $ x + 3 \ge 0 $, so the domain is $ x \ge -3 $. Since $ \sqrt{x+3} \ge 0 $, we have $ -2\sqrt{x+3} \le 0 $, hence $ g(x) \le 1 $: the range is $ y \le 1 $.

b) $ g(1) = -2\sqrt{4} + 1 = -3 $. For $ g(x) = -5 $: $ -2\sqrt{x+3} = -6 $, so $ \sqrt{x+3} = 3 $ and $ x + 3 = 9 $, giving $ x = 6 $. Check: $ g(6) = -2(3) + 1 = -5 $.

c) The parent $ \sqrt{x} $ is shifted 3 units left, stretched vertically by 2, reflected over the x-axis and shifted 1 unit up. The curve starts at $ (-3, 1) $ and falls, as in the figure.

16 Inverse with a restricted domain ★★★

a) Without a restriction the parabola fails the horizontal line test (for instance $ f(2) = f(4) = 2 $), so it has no inverse function. For $ x \ge 3 $ the function is one-to-one.

b) Swap: $ x = (y-3)^2 + 1 $, so $ (y-3)^2 = x - 1 $. Because $ y \ge 3 $, take the positive root: $ y - 3 = \sqrt{x - 1} $. Thus $ f^{-1}(x) = \sqrt{x-1} + 3 $. Its domain is the range of $ f $, namely $ x \ge 1 $; its range is the domain of $ f $, namely $ y \ge 3 $.

c) $ f(5) = 4 + 1 = 5 $ and $ f^{-1}(5) = \sqrt{4} + 3 = 5 $. The point $ (5, 5) $ lies on $ y = x $, so it is its own mirror image.

17 Does the order matter? ★★★

a) $ f(g(x)) = \sqrt{x - 4} $ and $ g(f(x)) = \sqrt{x} - 4 $.

b) For $ f(g(x)) $ we need $ x - 4 \ge 0 $: domain $ x \ge 4 $. For $ g(f(x)) $ we need $ x \ge 0 $: domain $ x \ge 0 $.

c) They are different functions, with different domains. For example $ f(g(5)) = \sqrt{1} = 1 $ but $ g(f(5)) = \sqrt{5} - 4 \approx -1.76 $; and $ x = 1 $ is allowed in the second but not in the first.

18 Parking fees ★★★

a) $ P(1.5) = 4 $. $ P(5) = 4 + 3(3) = 13 $. $ P(12) = 25 $.

b) $ 4 + 3(t-2) = 19 $ gives $ 3(t-2) = 15 $, so $ t = 7 $ hours.

c) At $ t = 9 $: $ 4 + 3(7) = 25 $, exactly the cap, so the pieces connect at $ (9, 25) $. The graph is a flat segment, a rising segment, then a flat segment again.

19 Absolute value as a piecewise function ★★★

a) If $ x \ge 1 $, then $ |x-1| = x - 1 $ and $ h(x) = 2x - 2 - 3 = 2x - 5 $. If $ x \lt 1 $, then $ |x-1| = 1 - x $ and $ h(x) = 2 - 2x - 3 = -2x - 1 $.

b) $ 2|x-1| = 3 $ gives $ |x-1| = 1.5 $, so $ x = 2.5 $ or $ x = -0.5 $. Check: $ h(2.5) = 2(1.5) - 3 = 0 $.

c) The vertex is $ (1, -3) $, the graph opens upward, so the range is $ y \ge -3 $.

20 Discount or coupon first? ★★★

a) Discount first: $ f(60) = 48 $, then $ g(48) = 43 $. Coupon first: $ g(60) = 55 $, then $ f(55) = 44 $.

b) $ g(f(x)) = 0.8x - 5 $ and $ f(g(x)) = 0.8(x-5) = 0.8x - 4 $. Applying the discount first is always cheaper, by exactly \$1, whatever the price (as long as the price is large enough to use the coupon).

21 Transforming a table of values ★★★

a) The input shift $ x + 1 $ moves each point 1 unit left; the outputs are doubled, then lowered by 3. The point $ (a, b) $ becomes $ (a - 1,\ 2b - 3) $: $ (-2, 1) \to (-3, -1) $; $ (0, 3) \to (-1, 3) $; $ (1, -2) \to (0, -7) $; $ (4, 0) \to (3, -3) $.

b) $ f(2x) $ compresses the graph horizontally by the factor $ \tfrac12 $: $ (a, b) \to (a/2, b) $: $ (-1, 1) $, $ (0, 3) $, $ (0.5, -2) $, $ (2, 0) $.

22 Proving two functions are inverses ★★★

$ f(g(x)) = \dfrac{2\cdot\frac{3x+1}{x-2} + 1}{\frac{3x+1}{x-2} - 3} $. Multiply the numerator and denominator by $ x - 2 $:

numerator: $ 2(3x+1) + (x-2) = 7x $; denominator: $ (3x+1) - 3(x-2) = 7 $.

So $ f(g(x)) = \dfrac{7x}{7} = x $ (for $ x \ne 2 $). This shows that $ f $ undoes $ g $. To be fully sure that they are inverses, also verify $ g(f(x)) = x $: $ \dfrac{3(2x+1) + (x-3)}{(2x+1) - 2(x-3)} = \dfrac{7x}{7} = x $. Both hold, so $ g = f^{-1} $.