Honors Algebra 2: Function Review and Challenge Questions (written by A. Beckwith)

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

1. Find all values of $$k$$ that make the following relation a non-function
{(2, 5), (-3, 11), ($$k$$, 5), ($$2k + 3$$, 5), (-7, 8)}
2. Solution {-3, -5, -7}
3. Given $$f(x) = -x^2 + 2x + 1$$ and $$g(x) = 2x + 1$$, find...
1. $$g^{-1} \left( f(-4) \right)$$
2. Solution -12
3. $$f \left( g(x) \right)$$
4. Solution $$-4x^2 + 2$$
5. $$f(-x)$$
6. Solution $$-x^2 - 2x + 1$$
4. Given that $$f(x)$$ has a domain $$\{ x \in \Re \mid -3 < x < 10 \}$$ and a range of {-1, 0, 1}
and $$g(x)$$ has a domain $$\{ x \in \Re \mid -1 \leq x \leq 1 \}$$ and a range of $$\{ y \in \Re \mid -3 \leq y < 0 \}$$
and $$h(x)$$ has a domain $$\{ x \in \Re \mid 50 < x < 100 \}$$ and a range of $$\{ y \in \Re \mid -3 < y < 10 \}$$ find...
1. the domain of $$g^{-1}(x)$$
2. Solution $$\{ x \in \Re \mid -3 \leq x < 0 \}$$
3. the range of $$f(3.1x - 9999)$$
4. Solution {-1, 0, 1}
5. The domain and range of the following. If either can't be determined, so state.
1. f(g(x))
2. Solution domain: $$\{ x \in \Re \mid -1 \leq x \leq 1 \}$$; range: can't be determined
3. $$g \left( f(x) \right)$$
4. Solution domain: $$\{ x \in \Re \mid -3 < x < 10 \}$$; range: can't be determined
5. $$h \left( f(x) \right)$$
6. Solution domain: $$\emptyset$$; range: $$\emptyset$$
7. $$f \left( h(x) \right)$$
8. Solution domain: $$\{ x \in \Re \mid 50 < x < 100 \}$$; range: {-1, 0, 1}
9. $$f \left( f(x) \right)$$
10. Solution domain: $$\{ x \in \Re \mid -3 < x < 10 \}$$; range: can't be determined
11. $$f \left( f(-x) \right)$$
12. Solution domain: $$\{ x \in \Re \mid -10 < x < 3 \}$$; range: can't be determined
5. For some function $$f(x)$$ the following is true: $$f(x) + 2 f (3 - x) = 4x + 5$$. Find $$f(1)$$.
6. Solution Plug in $$x = 1: f(1) + 2f(2) = 9$$. Now plug in $$x = 2: f(2) + 2f(1) = 13$$. You now have a system of 2 equations in two unknowns you can solve. $$\displaystyle f(1) = \frac{17}{3}$$
7. If $$f(x) = ax + b$$, find all values of $$a$$ and $$b$$ such that $$f \left( f \left( f(1) \right) \right) = 29$$ and $$f \left( f \left( f(0) \right) \right) = 2$$.
8. Solution $$f(1) = a + b$$; $$f(a + b) = a^2 + ab + b$$; $$f(a^2 + ab + b) = a^3 + a^2b + ab + b$$, so $$a^3 + a^2b + ab + b = 29$$. $$f(0) = b$$; $$f(b) = ab + b$$; $$f(ab + b) = a^2 + ab$$, so $$a^2b + ab = 2$$. Then $$a^3 + 2 = 29$$; $$a = 3$$; $$\displaystyle b = \frac{2}{13}$$; $$\displaystyle f(x) = 3x + \frac{2}{13}$$
9. Assume for some function $$f(x)$$ that $$f(x) = 1 - f(x - 1)$$ for all positive values of $$x$$. If $$f(2) = 12$$, find $$f(2012)$$.
10. Solution $$f(x + 1) = 1 - f(x) = 1 - (1 - f(x - 1)) = f(x - 1)$$. If $$f(x + 1) = f(x - 1)$$, then $$f(4) = f(2)$$ and $$f(6) = f(4)$$ and $$f(8) = f(6)$$, etc. Since $$f(2) = 12$$, $$f(2010) = 12$$.