Honors Pre-Calculus Unit 1 Review: Functions

A function is a rule that gives each input exactly one output. In $f(x) = 3x - 5$, the letter $f$ names the rule, $x$ is the input, and $f(x)$ is the output, so $f(4) = 3(4) - 5 = 7$. The notation $f(x)$ does not mean f times x.

The same function can show up as an equation, a table, a graph, or a sentence. "A plumber charges 20 dollars to show up plus 15 dollars per hour" is the verbal version of $C(h) = 15h + 20$. Expect to translate between these forms.

To check a graph, use the vertical line test. If any vertical line hits the graph twice, one input has two outputs, so the graph is not a function. A circle fails the test, and an upward-opening parabola passes.

Functions can be one-to-one, where no output repeats (like $2x + 1$), or many-to-one, where outputs repeat (like $x^2$, since $f(2) = f(-2) = 4$). Both count as functions. The difference matters in 1.7, because only one-to-one functions have inverses.

When a problem says "interpret," give the number with its units in context. If $P(6) = 500$, where $P$ is profit in dollars and the input is months, the answer is "the profit after 6 months is 500 dollars."

The domain is every input a function accepts, and the range is every output it can produce. Start by assuming all real numbers work, then remove anything that breaks the math.

Two restrictions cause most of the trouble. You cannot divide by zero, so for $h(x) = \frac{3}{x - 4}$ you exclude $x = 4$, and the domain is $(-\infty, 4) \cup (4, \infty)$. You also cannot take an even root of a negative number, so for $f(x) = \sqrt{2x + 6}$ you solve $2x + 6 \ge 0$ and get $[-3, \infty)$. Cube roots accept every real number.

Context can shrink the domain further. If $x$ counts people, only whole numbers make sense, even if the formula works for $x = 2.5$.

In interval notation, parentheses mean the endpoint is left out and brackets mean it is included. Infinity always gets a parenthesis because you never reach it. Set-builder notation says the same thing with a condition, such as $\{x \mid x \ge -3\}$.

Piecewise functions use different rules on different intervals. Graph each piece only on its own interval, with a closed dot for $\le$ or $\ge$ and an open dot for $<$ or $>$. For each piece, keep only the inputs in its stated interval where its formula is defined. The domain is the union of those inputs, and the range is the union of the outputs each piece actually produces.

The average rate of change of $f$ on $[a, b]$ is $\frac{f(b) - f(a)}{b - a}$. It is the slope of the secant line through the two endpoints, and it tells you the average change in output per one unit of input.

Worked example. Find the average rate of change of $f(x) = x^2$ on $[1, 4]$. First, $f(4) = 16$ and $f(1) = 1$. The change in output is $16 - 1 = 15$, and the change in input is $4 - 1 = 3$. So the rate is $\frac{15}{3} = 5$, meaning the output rises 5 units per unit of $x$ on average.

A function is increasing on an interval if its outputs rise as you move left to right, and decreasing if they fall. Always describe these intervals with $x$-values, not $y$-values. For example, $f(x) = -(x - 3)^2$ increases on $(-\infty, 3)$ and decreases on $(3, \infty)$.

A local maximum is an output that is at least as large as nearby outputs, like the height of a hilltop. An absolute maximum is the greatest output on the entire domain, and it can occur at more than one input. On a closed interval, check the endpoints too, because the absolute max or min can occur there.

When average rates of change approach a single limit as the interval shrinks around a point, that limit is the instantaneous rate of change. The limit does not exist at every point. This limit idea is part of calculus.

You can combine two functions by adding, subtracting, multiplying, or dividing their outputs at the same $x$. For a quotient $\left(\frac{f}{g}\right)(x)$, use inputs in the domains of both functions and remove any $x$ where $g(x) = 0$.

Composition works differently. In $(f \circ g)(x) = f(g(x))$, the output of $g$ becomes the input of $f$. Work from the inside out, so $g$ goes first even though $f$ is written first. Order matters. With $f(x) = 2x + 1$ and $g(x) = x^2$, you get $f(g(x)) = 2x^2 + 1$ but $g(f(x)) = (2x + 1)^2$.

Worked example. Let $f(x) = \sqrt{x}$ and $g(x) = x^2 - 4$. Find $(f \circ g)(3)$ and the domain of $f \circ g$. Inside first, $g(3) = 9 - 4 = 5$, so $(f \circ g)(3) = \sqrt{5}$. For the domain, $x$ must work in $g$ (all reals do) and $g(x)$ must work in $f$, so $x^2 - 4 \ge 0$. That gives $x \le -2$ or $x \ge 2$, so the domain is $(-\infty, -2] \cup [2, \infty)$.

Decomposition runs this backward. For $h(x) = \sqrt{2x - 1}$, the outer function is $f(x) = \sqrt{x}$ and the inner one is $g(x) = 2x - 1$. More than one split can work, but the simplest one is usually what you want. You will use this move for the chain rule in calculus.

Transformations let you start from a parent graph you know, like $y = x^2$ or $y = \sqrt{x}$, and move it instead of graphing from scratch. Changes outside the function act on $y$ the way they look. Changes inside act on $x$ in the opposite way.

Outside the function, $f(x) + k$ shifts the graph vertically by $k$. For nonzero $a$, $a \cdot f(x)$ scales the graph vertically by $|a|$: it stretches when $|a| > 1$ and compresses when $0 < |a| < 1$. A negative $a$ also reflects the graph over the $x$-axis. If $a = 0$, every output becomes zero. Inside the function, $f(x - h)$ shifts right $h$, so $f(x + 4)$ shifts left 4. The graph of $f(2x)$ is squeezed to half its width, and $f(-x)$ reflects over the $y$-axis.

Worked example. Describe $g(x) = -2(x - 3)^2 + 1$ starting from $f(x) = x^2$. Shift right 3, stretch vertically by 2, reflect over the $x$-axis, then shift up 1. The vertex moves from $(0, 0)$ to $(3, 1)$, and the parabola now opens down. To check one point, $(1, 1)$ on $f$ moves to $(4, -2 \cdot 1 + 1) = (4, -1)$, and $g(4) = -2(1)^2 + 1 = -1$.

Symmetry connects here. An even function has a domain symmetric about zero and satisfies $f(-x) = f(x)$, so its graph is symmetric about the $y$-axis, like $x^2$ on the real numbers. An odd function has a domain symmetric about zero and satisfies $f(-x) = -f(x)$, so its graph is symmetric about the origin, like $x^3$ on the real numbers. Most functions, such as $x^2 + x$, are neither.

The parent function $f(x) = |x|$ is a V with its vertex at the origin. It is really a piecewise function, equal to $x$ when $x \ge 0$ and $-x$ when $x < 0$. That is why each arm is a straight line, with slope 1 on the right and $-1$ on the left.

The form $f(x) = a|x - h| + k$ follows the transformation rules from 1.5. If $a \ne 0$, the vertex is $(h, k)$ and the axis of symmetry is $x = h$. If $a > 0$, the V opens up and the range is $y \ge k$. If $a < 0$, it opens down and the range is $y \le k$. If $a = 0$, the graph is the horizontal line $y = k$ and its range is $\{k\}$. The domain is all real numbers in each case.

Worked example. Solve $3|2x - 1| + 7 = 22$. Isolate the absolute value first by subtracting 7 and dividing by 3, which gives $|2x - 1| = 5$. The inside is either 5 or $-5$. From $2x - 1 = 5$ you get $x = 3$, and from $2x - 1 = -5$ you get $x = -2$. Both values check in the original equation.

If an isolated absolute value equals a negative number, as in $|x + 1| = -4$, there is no solution. Absolute value measures distance, so it is never negative. The same distance idea explains why $|a - b|$ is the gap between $a$ and $b$ on a number line, and why a tolerance like $|x - 10| \le 0.2$ means within 0.2 of 10.

An inverse function undoes the original. If $f(3) = 9$, then $f^{-1}(9) = 3$. Inputs and outputs trade places, so the domain of $f$ becomes the range of $f^{-1}$, and the range of $f$ becomes the domain of $f^{-1}$.

Only one-to-one functions have inverses, and the horizontal line test checks for this. The function $x^2$ fails, since $f(3) = f(-3) = 9$, so $f^{-1}(9)$ would have no single answer. Restricting the domain to $x \ge 0$ fixes the problem, and then $f^{-1}(x) = \sqrt{x}$.

Worked example. Find the inverse of $f(x) = 3x - 2$. Write $y = 3x - 2$, then swap $x$ and $y$ to get $x = 3y - 2$. Solving for $y$ gives $y = \frac{x + 2}{3}$, so $f^{-1}(x) = \frac{x + 2}{3}$. To verify, compose both ways. You get $f(f^{-1}(x)) = 3 \cdot \frac{x + 2}{3} - 2 = x$ and $f^{-1}(f(x)) = \frac{(3x - 2) + 2}{3} = x$. Both compositions must equal $x$.

On a graph, the inverse is the reflection of $f$ over the line $y = x$. Every point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$. This is exactly how $y = 2^x$ and $y = \log_2 x$ are related, which you will use in the exponential and logarithm unit.