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
Find all values of \(k\) that make the following relation a non-function
{(2, 5), (-3, 11), (\(k\), 5), (\(2k + 3\), 5), (-7, 8)}
Solution
{-3, -5, -7}
Given \(f(x) = -x^2 + 2x + 1\) and \(g(x) = 2x + 1\),
find...
\( g^{-1} \left( f(-4) \right) \)
Solution
-12
\( f \left( g(x) \right) \)
Solution
\(-4x^2 + 2\)
\( f(-x) \)
Solution
\( -x^2 - 2x + 1 \)
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...
the domain of \(g^{-1}(x)\)
Solution
\( \{ x \in \Re \mid -3 \leq x < 0 \} \)
the range of \(f(3.1x - 9999)\)
Solution
{-1, 0, 1}
The domain and range of the following. If either can't be determined, so state.
f(g(x))
Solution
domain: \( \{ x \in \Re \mid -1 \leq x \leq 1 \} \); range: can't be determined
\( g \left( f(x) \right) \)
Solution
domain: \( \{ x \in \Re \mid -3 < x < 10 \} \); range: can't be determined
Solution
domain: \( \{ x \in \Re \mid 50 < x < 100 \} \); range: {-1, 0, 1}
\( f \left( f(x) \right) \)
Solution
domain: \( \{ x \in \Re \mid -3 < x < 10 \} \); range: can't be determined
\( f \left( f(-x) \right) \)
Solution
domain: \( \{ x \in \Re \mid -10 < x < 3 \} \); range: can't be determined
For some function \(f(x)\) the following is true: \( f(x) + 2 f (3 - x) = 4x + 5 \). Find \( f(1)\).
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. \( \displaystyle f(1) = \frac{17}{3} \)
If \(f(x) = ax + b\), find all values of \(a\) and \(b\) such that \( f \left( f \left( f(1) \right) \right) = 29\) and \( f \left( f \left( f(0) \right) \right) = 2\).
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 \(a^3 + 2 = 29\); \(a = 3\); \( \displaystyle b = \frac{2}{13}\); \( \displaystyle f(x) = 3x + \frac{2}{13}\)
Assume for some function \(f(x)\) that \(f(x) = 1 - f(x - 1)\) for all positive values of \(x\). If \(f(2) = 12\), find \(f(2012)\).
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 \(f(2) = 12\), \(f(2010) = 12\).