College Algebra & Precalculus Bridge: lessons and practice

Read the prerequisite, study a worked example and justify each step in these selected university topics.

STAGE UPRE 1.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Functions, Domains & Transformations

Useful preparation: Real Function Domains

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For a long equation, swipe sideways inside its formula box to see every term.

Meaning & notation

A function combines a rule with permitted inputs. Before calculating, identify possible inputs, attained outputs, and whether distinct inputs can share an output. These questions control composition, inverses, transformations, and later calculus.

Domain and range: The domain consists of permitted inputs; the range consists of outputs actually attained. An algebraic formula may have a natural real domain, while a model can impose additional restrictions.
Composition: Composition applies the inner function first, then sends its output into the outer function. The intermediate output must belong to the outer domain.
Inverse function: An inverse reverses input and output on specified sets. It exists as a function on the range only when the original function is one-to-one.

Conditions before calculation

Work over the real numbers here: denominators cannot vanish, even-root radicands must be nonnegative, and logarithm arguments must be positive.
Record exclusions before simplifying. Canceling a factor does not automatically put an excluded input back into the original domain.
An inverse requires a one-to-one domain restriction when necessary; its domain is the original range, not automatically all real numbers.

Why the method works

1. Translate each operation into a restriction

Intersect every restriction. In a composition, both the initial input and the intermediate output must be permitted.

D_{f\circ g}=\{x\in D_g:g(x)\in D_f\}

2. Track transformations through coordinates

Horizontal changes act on inputs, so solve backward for the new coordinate. For nonzero b, graph points transform as shown; vertical changes act directly.

g(x)=a f(b(x-h))+k,\qquad (u,f(u))\mapsto\left(h+\frac{u}{b},\,a f(u)+k\right)

3. Distinguish holes from asymptotes

An excluded input may create a finite hole or unbounded behavior. Limits distinguish them; a zero denominator alone does not.

\frac{x^2-9}{x-3}=x+3\quad(x\ne3),\qquad \lim_{x\to3}\frac{x^2-9}{x-3}=6

Worked examples

EXAMPLE 1

Determine the domain, zero, and vertical asymptote of the displayed function.

Impose the square-root restriction and denominator exclusion simultaneously.
$f(x)=\frac{\sqrt{x-2}}{x-5},\qquad x\ge2,\quad x\ne5$
The numerator vanishes at the allowed left endpoint. Near the excluded point it instead approaches a positive number, so cancellation cannot remove the singularity.
$D_f=[2,5)\cup(5,\infty),\qquad f(2)=0,\qquad \sqrt{x-2}\to\sqrt3\ (x\to5)$
The denominator changes sign through zero while the numerator stays positive. This establishes the asymptote through one-sided behavior.
$\lim_{x\to5^-}f(x)=-\infty,\qquad\lim_{x\to5^+}f(x)=+\infty$

Conclusion: The zero is 2 and the vertical asymptote is the line with horizontal coordinate 5.

EXAMPLE 2

Find an invertible restriction of the shifted quadratic and determine the inverse.

On the whole real line, inputs equally far from 1 have the same output. Restrict to the right half, where distance from 1 increases.
$q(x)=(x-1)^2+2,\qquad x\in[1,\infty)$
Solve for the input. The restriction selects the nonnegative square-root branch rather than both algebraic signs.
$y-2=(x-1)^2,\qquad x=1+\sqrt{y-2}$
Exchange input and output names and check composition on the original domain. The range supplies the inverse domain.
$q^{-1}(y)=1+\sqrt{y-2},\quad y\ge2;\qquad q^{-1}(q(x))=1+|x-1|=x\quad(x\ge1)$

Conclusion: The restriction has range starting at 2 and an inverse defined exactly on that range.

Pitfalls & counterexamples

Tempting claim: Algebraically equal expressions always define the same function.

Correction: Functions must also have the same domain and matching outputs at every input in that domain.

Test the claim: The canceled rational expression in the hole example excludes 3, while the linear expression alone includes 3.

Check your understanding

Can the unrestricted squaring function have an inverse on the real numbers?

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Compare your reasoning

No.

A positive number and its negative have the same square, so an output would not identify a unique original input.

Connect this idea

Domain tracking prepares limit arguments; one-to-one restrictions explain inverse trigonometric functions; transformations anticipate the chain rule.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

Preview the eight practice prompts

Problem 1
For $\displaystyle f\, \left(x\right) = \frac{1}{x - 2}$ , what input is excluded?
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Problem 2
Let $\displaystyle f\, \left(x\right) = \frac{2}{x - 1} + 3$ . What is its horizontal asymptote y-value?
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Problem 3
Solve $\displaystyle \operatorname{log}_{2}\, \left(x\right) = 2$ . Find x.
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Problem 4
For $\displaystyle y = \operatorname{sin}\, \left(2\, x\right)$ , the period is $\displaystyle a\, \pi$ . Find a.
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Problem 5New context
A reciprocal response is modeled by $\displaystyle f\, \left(x\right) = \frac{1}{x - 3}$ , what input is excluded?
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Problem 6New context
A transformed response is $\displaystyle f\, \left(x\right) = \frac{3}{x - 1} + 4$ . What is its horizontal asymptote y-value?
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Problem 7New context
A logarithmic calibration requires $\displaystyle \operatorname{log}_{2}\, \left(x\right) = 3$ . Find x.
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Problem 8New context
An oscillator follows $\displaystyle y = \operatorname{sin}\, \left(3\, x\right)$ , the period is $\displaystyle a\, \pi$ . Find a.
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