Wemazi

Topic 1.12

Find where an expression is defined

Builds on explaining why continuity fails at one point. 6 minutes.

A curve that exists on only one side

The curve is the graph of \(f(x) = \sqrt{x - 1}\), and the orange dot is on it at the input \(x\) that you choose with the slider.

Set the input to 5. The expression under the root is \(5 - 1 = 4\), and \(\sqrt{4} = 2\), so the dot is at height 2.

Set the input to 1. The expression under the root is 0, and \(\sqrt{0} = 0\), so the dot is at the start of the curve.

Set the input below 1, for example to 0. The expression under the root is \(-1\), and no real number squared gives \(-1\), so the function has no value there. The dot disappears, and a red ring marks that input on the axis.

The inputs where the function has a value form its domain. Press Show the allowed inputs, and the teal bar on the axis covers the domain, \(x \ge 1\).

Three restrictions

The numbers are gone. Press a button to switch between three expressions. In each, \(u\) stands for the expression inside, and the dashed line marks the input \(c\) where \(u = 0\).

A fraction \(\frac{1}{u}\) has no value where \(u = 0\), because division by 0 has no meaning. The graph has a gap at \(c\).

A square root \(\sqrt{u}\) has no real value where \(u\) is negative, so the graph has no points on that side of \(c\). A red bar on the axis marks those inputs. A logarithm \(\ln u\) has no value where \(u\) is 0 or negative, so its graph has no points there either.

Notation

  1. To find the domain, list the inputs that the formula cannot accept.

  2. A denominator cannot be 0.

    \[\frac{1}{u}: \quad u \ne 0\]

  3. The radicand, the expression under a square root, cannot be negative.

    \[\sqrt{u}: \quad u \ge 0\]

  4. A logarithm needs a positive input. The number \(\ln u\) is the power of \(e\) that gives \(u\), and a power of \(e\) is never 0 or negative.

    \[\ln u: \quad u > 0\]

  5. A function built from these parts is continuous at every input of its domain.

When it fails

Press Test the edge input \(c\), where \(u = 0\). The fraction and the logarithm have no value there, so \(c\) is not in their domain.

The square root has the value \(\sqrt{0} = 0\) at \(c\), so \(c\) is in its domain. Test each edge input on its own.

Example 1: A denominator

Find the domain of \(f(x) = \dfrac{7}{x^2 - x - 12}\).

  1. A denominator cannot be 0. Find the inputs that make it 0.

    \[x^2 - x - 12 = 0\]

  2. Factor the left side. A product is 0 only when a factor is 0, so \(x = 4\) or \(x = -3\). The red rings mark these inputs.

    \[(x - 4)(x + 3) = 0\]

  3. The curve breaks at each of these inputs, and the teal bars cover every other input. The domain is all real numbers except \(-3\) and 4.

    \[x \ne -3, \quad x \ne 4\]

Example 2: A logarithm

Find the domain of \(g(x) = \ln(6 - 2x)\).

  1. A logarithm needs a positive input, so \(6 - 2x\) must be greater than 0.

    \[6 - 2x > 0\]

  2. A tempting wrong move: Subtract 6 and divide both sides by \(-2\) without reversing the inequality. This gives \(x > 3\).

    \[x > 3\]

    Test \(x = 4\), which satisfies \(x > 3\). Then \(6 - 2(4) = -2\), and \(\ln(-2)\) has no value, so the curve has no point above the red ring.

  3. Subtracting 6 gives \(-2x > -6\). Dividing by a negative number reverses the inequality, so this gives \(x < 3\).

    \[\begin{aligned} 6 - 2x &> 0 \\ -2x &> -6 \\ x &< 3 \end{aligned}\]

  4. At \(x = 3\), the logarithm receives \(6 - 2(3) = 0\), so 3 is not allowed either. The domain is \(x < 3\), and the curve falls without bound as \(x\) approaches 3.

Practice

Find the domain of \(p(x) = \sqrt{x^2 - 1}\).

The root has a value where the radicand \(x^2 - 1\) is not negative. Move the slider to see which inputs have a value. Then find the two inputs where the radicand is 0.

Solution

  1. The radicand must not be negative.

    \[x^2 - 1 \ge 0\]

  2. The radicand is 0 at \(x = -1\) and \(x = 1\), because \(x^2 - 1 = (x - 1)(x + 1)\).

  3. These two inputs split the x-axis into three intervals, so test one input in each. At \(x = -2\), the radicand is 3, which is positive. At \(x = 0\), it is \(-1\), which is negative. At \(x = 2\), it is 3.

  4. The domain is \(x \le -1\) or \(x \ge 1\). The edges belong to it, because \(\sqrt{0} = 0\).

Questions

Use the graph of \(f(x) = \ln(x - 1)\) for the first two questions. The faint dashed line is at \(x = 1\).

  1. Which inputs belong to the domain of \(f\)?

    • \(x > 1\)
    • \(x \ge 1\)
    • \(x < 1\)
    • All real numbers
    Answer

    \(x > 1\)

  2. Which input has no point on the curve?

    • \(x = 0\)
    • \(x = 1.5\)
    • \(x = 3\)
    • \(x = 10\)
    Answer

    \(x = 0\)

  3. Find the domain of \(q(x) = \dfrac{1}{x^2 - 5x}\).

    • \(x \ne 0\) and \(x \ne 5\)
    • \(x \ne 5\) only
    • \(x \ne 5\) and \(x \ne -5\)
    • \(x > 5\)
    Answer

    \(x \ne 0\) and \(x \ne 5\)

Summary

A denominator cannot be 0, a radicand cannot be negative, and a logarithm needs a positive input.

The inputs that remain form the domain, which for \(\sqrt{x - 1}\) is the teal bar \(x \ge 1\).

These parts have no break where they have a value, so the function is also continuous at every input of its domain.

This is one of four skills in topic 1.12 of Which & Why: AP Calculus.