Functions, Patterns and Sequences flashcards

Each input has exactly one output. {(1, 3), (2, 3)} is a function, since inputs may share an output; {(4, 1), (4, 7)} is not, since input 4 has two outputs. On a graph, use the vertical line test.

f(x) names the output at input x; it is not f times x. f(3) = 2(3) − 5 = 1. To find the input for a given output, write rule = output and solve for x.

Domain: allowed inputs. Range: outputs produced. Exclude inputs that make a denominator zero ($\frac{1}{x-5}$ excludes 5) or a square root's inside negative ($\sqrt{x+1}$ needs x ≥ −1).

$m=\frac{y_2-y_1}{x_2-x_1}$, change in output over change in input. From (2, 5) to (6, 17): m = 12/4 = 3. In f(x) = mx + b, b is the output when x = 0.

$a_n=a_1+(n-1)d$, where $a_1$ is the first term and d the common difference. For 7, 11, 15, ...: $a_{10}=7+9(4)=43$. Using n instead of n − 1 overshoots by d.

$S_n=\frac{n(a_1+a_n)}{2}$: the average of the first and last terms times the number of terms. 3, 7, 11, ..., 23 has 6 terms, so the sum is 6(26)/2 = 78.

$a_n=a_1\cdot r^{n-1}$, where r is the common ratio (any term divided by the one before). For 3, 6, 12, ...: $a_6=3\cdot 2^5=96$. Growth has r > 1; decay has 0 < r < 1.

Work inside out: compute g(x), then feed that result into f. With $f(x)=x^2$ and g(x) = x + 1: f(g(2)) = f(3) = 9, but g(f(2)) = g(4) = 5, so order matters.