1.5 Functions, Graphs, Transformations, and Inverses

1.5 Functions, Graphs, Transformations, and Inverses

Calculus questions often become elementary once the domain, composition order, or graph transformation is read correctly. A function $f:A\to B$ assigns one value $f(x)\in B$ to every $x\in A$. The set $A$ is the domain, the set of attained outputs is the range, and the codomain $B$ need not equal the range.

Domains and Algebraic Families

For a formula over the real numbers, impose every restriction simultaneously. Denominators cannot vanish, even-index radicals require nonnegative radicands, and logarithm arguments must be positive. Thus $f(x)=\frac{\sqrt{2x-1}}{\log(x-2)}$ has $x\ge 1/2$, $x>2$, and $x\ne3$, so its domain is $(2,3)\cup(3,\infty)$. A polynomial has all real numbers as its domain. A rational function $p/q$ is defined where $q\ne0$; canceled factors still mark holes in the original function.

Parity and symmetry save time. If $f(-x)=f(x)$, $f$ is even and its graph is symmetric about the $y$-axis. If $f(-x)=-f(x)$, it is odd and has origin symmetry. Periodicity means $f(x+T)=f(x)$ for some $T>0$.

Composition and Piecewise Functions

The composition $(f\circ g)(x)=f(g(x))$ is defined only when $x$ lies in the domain of $g$ and $g(x)$ lies in the domain of $f$. Composition is generally not commutative. For $f(x)=\sqrt{x}$ and $g(x)=x-2$, $f\circ g=\sqrt{x-2}$ has domain $[2,\infty)$, while $g\circ f=\sqrt{x}-2$ has domain $[0,\infty)$.

For piecewise functions, use the formula assigned to the input. Continuity at a breakpoint $c$ requires the left limit, right limit, and assigned value to agree. Differentiability further requires matching one-sided derivatives. Absolute-value equations can be handled piecewise: $|u|=a$ with $a\ge0$ means $u=a$ or $u=-a$, while $|u|<a$ means $-a<u<a$.

Graph Transformations

Starting from $y=f(x)$:

• $f(x-h)+k$ shifts right by $h$ and up by $k$.
• $a f(x)$ stretches vertically by $|a|$ and reflects across the $x$-axis if $a<0$.
• $f(bx)$ compresses horizontally by $|b|$ when $|b|>1$ and reflects across the $y$-axis if $b<0$.
• $|f(x)|$ reflects portions below the axis upward; $f(|x|)$ copies the right half of the graph to the left.

The horizontal rule is inverse: $f(2x-6)=f(2(x-3))$ is shifted right 3 and compressed horizontally by factor $1/2$.

One-to-One Functions and Inverses

A function is one-to-one if $f(x_1)=f(x_2)$ implies $x_1=x_2$. Graphically this is the horizontal-line test. A bijection has an inverse $f^{-1}$ satisfying $f^{-1}(f(x))=x$ on the original domain and $f(f^{-1}(y))=y$ on the original range. The graph of the inverse is the reflection of the graph across $y=x$.

To find an inverse, write $y=f(x)$, solve for $x$, and interchange variables while carrying the domain and range. The squaring function is not one-to-one on all of $\mathbb R$, but its restriction $f:[0,\infty)\to[0,\infty)$ has inverse $f^{-1}(x)=\sqrt{x}$. Do not confuse $f^{-1}(x)$ with $1/f(x)$.

Worked example

Let $f(x)=(3x-2)/(x+4)$. Solving $y=(3x-2)/(x+4)$ gives $yx+4y=3x-2$, hence $x(y-3)=-2-4y$ and $f^{-1}(y)=\frac{-2-4y}{y-3}=\frac{4y+2}{3-y}.$ The original domain excludes $-4$ and its range excludes $3$; the inverse domain therefore excludes $3$ and its range excludes $-4$.

Common traps

A canceled denominator factor removes an algebraic factor but does not restore an excluded input. An inverse swaps domain and range, not numerator and denominator. For compositions, check both domain stages. When a graph is transformed, handle the inside expression first and remember that horizontal scale and direction are reversed.

Exponential, logarithmic, and rational structure

Exponential and logarithmic functions are inverse families, so $a^{\log_a x}=x$ for $x>0$ and $\log_a(a^x)=x$ for every real $x$, where $a>0$ and $a\ne1$. Their transformations follow the same inside-out rules as other graphs. For example, $2^{x-3}+1$ shifts $2^x$ right 3 and up 1, while $\log_2(x+4)$ shifts $\log_2x$ left 4 and has vertical asymptote $x=-4$. A logarithmic equation must be checked against the original positive-argument restrictions after it is solved.

For a rational function $r(x)=p(x)/q(x)$, first factor both polynomials. A canceled factor produces a removable hole at the excluded input; an uncanceled zero of $q$ is a vertical-asymptote candidate. End behavior comes from degrees: if $\deg p<\deg q$, then $y=0$ is a horizontal asymptote; equal degrees give the ratio of leading coefficients; a numerator degree exactly one larger suggests a slant asymptote found by polynomial division.

Structural comparison

These facts often settle a domain, range, or asymptote question before any derivative is needed.