Nonlinear functions and function notation

This topic is about reading function notation and exponentials for what they actually say, rather than for what their symbols look like. A rule such as $f(x)=5x+2$ is an instruction: wherever an input appears in the parentheses, substitute it for $x$ and simplify. A composition $f(g(x))$ stacks two of those instructions and must be read from the inside out. An exponential equation such as $2^{x}=32$ asks a question about a power, and a model such as $y=3(2)^{x}$ packs a starting value, a per-step factor, and a step count into one short expression. The points leak in four predictable places: reading $f(x)$ as multiplication, composing in the wrong order, treating an exponent like a coefficient, and misnaming the parts of a growth model. Each cure is the same idea: slow down and read what the notation is asking before you compute.

When a function is named, like $f(x)=5x+2$, the symbol $f(x)$ is a label for the output, not a product. The parentheses do not mean $f$ times $x$; they hold the input. To evaluate $f(3)$, you substitute $3$ for every $x$ in the rule and simplify: $f(3)=5(3)+2=17$. The most common slip is to multiply the visible numbers instead of substituting, reading $f(3)$ as $2$ times $3$, or $5$ times $2$, or just $5(3)$ with the $+2$ forgotten. None of those is what the notation asks. Substitute the whole input into the whole rule.

• $f(x)$ is a name for the output; the parentheses hold the input, not a multiplication.
• To find $f(\text{input})$, replace every $x$ in the rule with that input.
• Simplify the entire rule, including every constant term, before reporting the value.

Worked example. If $f(x)=5x+2$, find $f(3)$.

A composition like $f(g(2))$ means apply $g$ to $2$ first, then apply $f$ to whatever $g$ produced. The inner function runs first because it sits inside the parentheses. Order matters: $f(g(2))$ and $g(f(2))$ are almost always different numbers, so swapping them is a real mistake, not a harmless rearrangement. The two other slips are stopping at the inner value without applying the outer function, and adding $f(2)$ and $g(2)$ instead of feeding one into the other. Evaluate the inside, then carry that single number outward.

• In $f(g(x))$, evaluate the inner function $g$ first, then apply $f$ to its output.
• $f(g(x))$ and $g(f(x))$ are different in general, so the order cannot be swapped.
• Carry the inner output forward as the new input; do not stop at it and do not add the two outputs.

Worked example. If $f(x)=3x+1$ and $g(x)=2x+5$, find $f(g(2))$.

An equation like $2^{x}=32$ has the variable in the exponent, and that changes what you are solving for. You are not undoing a coefficient; you are asking a question about a power: $2$ raised to what power equals $32$? Since $2^{5}=32$, the answer is $x=5$. The tempting wrong move is to treat the exponent like a coefficient and divide, writing $x=32 \div 2=16$, which is linearizing an equation that is not linear. Rewriting both sides as powers of the same base makes the exponent readable directly. The same idea handles negative exponents: $2^{x}=\dfrac{1}{8}$ asks for the power of $2$ that gives a reciprocal, and since $\dfrac{1}{8}=2^{-3}$, the exponent is $-3$.

• In $b^{x}=N$, the unknown is the exponent; ask what power of $b$ gives $N$.
• Do not divide $N$ by $b$. The exponent is not a coefficient, and dividing linearizes the equation.
• Rewrite $N$ as a power of $b$ and read the exponent. A reciprocal target means a negative exponent.

Worked example. Solve $2^{x}=32$, then $2^{x}=\dfrac{1}{8}$.

A model written as $y=a(b)^{x}$ carries three separate pieces of information, and most errors come from mixing them up. The coefficient $a$ is the initial value, the output when $x=0$. The base $b$ is the growth or decay factor, what you multiply by for each one-step increase in $x$. The exponent $x$ counts how many of those steps have happened. To find the output at a particular $x$, multiply the initial value by the factor that many times. Reporting the initial value as the answer, multiplying by the factor only once, or dropping the coefficient entirely are all symptoms of not naming the parts first.

• The coefficient $a$ is the value at $x=0$, not the answer for every $x$.
• The base $b$ is the per-step factor; the exponent says how many times to apply it.
• Evaluate $a(b)^{x}$ as a whole: keep the coefficient and raise the base to the full exponent.

Worked example. A quantity is modeled by $y=3(2)^{x}$. Find $y$ at $x=3$.

Read $f(x)$ as an instruction to substitute, not a product; compose from the inside out and never swap the order; in $b^{x}=N$ ask what power of $b$ gives $N$ instead of dividing; and in $y=a(b)^{x}$ name the initial value, the factor, and the step count before you compute. Read what the notation asks, then evaluate.

Four patterns recur on these questions. They are the same ones the diagnostic routes on.