Functions and Transformations: math test solutions, Grade 11 – download the PDF

1 Notation and domains / 3 pts

a) \( f(-2) = 2(4) + 2 + 3 = 13 \). (1 pt)

b) \( 2x - 7 \ge 0 \) gives \( x \ge 3.5 \). (1 pt)

c) \( x^2 - 16 = 0 \) for \( x = \pm 4 \), so the domain is all real numbers except 4 and \( -4 \). (1 pt)

2 Transformations / 4 pts

a) Shift 2 units left, vertical stretch by 3, reflection over the x-axis, shift 5 units up. (1 pt for the shifts, 1 pt for the stretch and reflection)

b) The vertex is \( (-2, 5) \). (1 pt)

c) The graph opens downward from height 5: the range is \( y \le 5 \). (0.5 pt)

d) \( g(1) = -3|3| + 5 = -9 + 5 = -4 \). (0.5 pt)

3 Combining functions / 3 pts

a) \( (f+g)(x) = x^2 + 1 + 2x - 3 = x^2 + 2x - 2 \). (1 pt)

b) \( f(2) = 5 \) and \( g(2) = 1 \), so \( (fg)(2) = 5 \). (1 pt)

c) \( f(x) = x^2 + 1 \ge 1 \), so it is never zero: the domain is all real numbers. (1 pt)

4 Composition / 4 pts

a) \( f(g(x)) = 3(x^2 - 1) + 2 = 3x^2 - 1 \). (1 pt)

b) \( g(f(x)) = (3x+2)^2 - 1 = 9x^2 + 12x + 3 \). (1 pt)

c) \( f(g(-2)) = 3(4) - 1 = 11 \). (1 pt)

d) \( 3x^2 - 1 = 47 \) gives \( x^2 = 16 \), so \( x = 4 \) or \( x = -4 \). (1 pt)

5 Inverse functions / 4 pts

a) Swap: \( x = (y-2)^3 + 4 \), so \( (y-2)^3 = x - 4 \) and \( y = \sqrt[3]{x - 4} + 2 \). (2 pts)

b) \( f^{-1}(-4) = \sqrt[3]{-8} + 2 = 0 \). Check: \( f(0) = (-2)^3 + 4 = -4 \). (1 pt)

c) \( f(f^{-1}(x)) = \left(\sqrt[3]{x-4} + 2 - 2\right)^3 + 4 = (x - 4) + 4 = x \). (1 pt)

6 Shipping cost / 2 pts

\( 1.2 \le 2 \): \( S(1.2) = 6 \). (1 pt)

\( 6 \gt 2 \): \( S(6) = 6 + 2.5(4) = 16 \). The shipping costs \$16. (1 pt)