3.5 Absolute Value and Square Root Function Families

3.5 Absolute Value and Square Root Function Families

The College Board names eleven function families you must recognize "symbolically, graphically or in tabular form." Two of them — absolute value and square root — are the ones students most often meet only as equations to solve, never as functions with their own shapes, domains, and ranges. Because the 30% Representations area can show you a V-shaped graph or a half-parabola and ask which formula produces it, this section treats both families in their own right.

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1. The Absolute Value Function $f(x) = |x|$

The parent function is defined piecewise:

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

          y
          |
       3  +   \         /
       2  +    \       /
       1  +     \     /
          |      \   /
     -----+-------\ /-------- x
         -3  -2  -1 0  1  2  3
              f(x) = |x|

The general form

$f(x) = a|x - h| + k$

• Vertex: $(h, k)$ — the corner moves with the shifts.
• $a > 0$ opens upward (minimum at the vertex); $a < 0$ opens downward (maximum at the vertex).
• $|a|$ controls steepness: $|a| > 1$ narrows the V, $0 < |a| < 1$ widens it.
• Range: $[k, \infty)$ when $a > 0$; $(-\infty, k]$ when $a < 0$.

Worked Example 3.5A: Reading a V-shaped graph

A graph is a V opening downward with vertex at $(2, 5)$ and passing through $(0, 1)$. Find its equation.

Solution:

1. Start from $f(x) = a|x - 2| + 5$.
2. Substitute the point $(0, 1)$: $1 = a|0 - 2| + 5 = 2a + 5$.
3. Solve: $2a = -4$, so $a = -2$.
4. The equation is $f(x) = -2|x - 2| + 5$, with range $(-\infty, 5]$.

The negative $a$ matches "opens downward" — a good self-check.

Worked Example 3.5B: Absolute value as a piecewise function

Rewrite $g(x) = |3x - 6|$ without absolute value bars.

Solution: The interior expression changes sign at $3x - 6 = 0$, i.e. $x = 2$.

$g(x) = \begin{cases} 3x - 6 & \text{if } x \ge 2 \\ 6 - 3x & \text{if } x < 2 \end{cases}$

This is exactly the reflection rule: the portion of $y = 3x - 6$ that lies below the $x$-axis is flipped above it.

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2. The Square Root Function $f(x) = \sqrt{x}$

The last row matters: $\sqrt{x}$ grows without bound but ever more slowly. Compare $\sqrt{4} = 2$, $\sqrt{100} = 10$, $\sqrt{10000} = 100$ — quadrupling the input only doubles the output.

The general form

$f(x) = a\sqrt{x - h} + k$

The endpoint of the curve sits at $(h, k)$. The graph extends rightward when the coefficient inside $x$ is positive and leftward when it is negative.

Finding the domain — the single most tested skill here

Set the radicand $\ge 0$ and solve.

Worked Example 3.5C: Domain and range of a transformed radical

Find the domain and range of $f(x) = -2\sqrt{8 - 2x} + 3$.

Solution:

1. Domain: require $8 - 2x \ge 0 \implies 8 \ge 2x \implies x \le 4$, giving $(-\infty, 4]$.
2. Endpoint: at $x = 4$ the radical is $0$, so $f(4) = 3$. The curve starts at $(4, 3)$ and extends leftward (because $x$ carries a negative coefficient inside).
3. Range: $\sqrt{8 - 2x} \ge 0$, so $-2\sqrt{8 - 2x} \le 0$, so $f(x) \le 3$. The range is $(-\infty, 3]$.

The square root and the squaring function are inverses — with a restriction

$f(x) = x^2$ is not one-to-one on $(-\infty, \infty)$, so it has no inverse there. Restrict its domain to $[0, \infty)$ and it becomes one-to-one, with inverse $f^{-1}(x) = \sqrt{x}$. This is why

$\sqrt{x^2} = |x|, \quad \text{not } x$

For $x = -5$: $\sqrt{(-5)^2} = \sqrt{25} = 5 = |-5|$. This identity is the bridge between the two families in this section, and it is a frequent distractor source.

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3. Comparing the Two Families

The last row is the practical test: a horizontal line meets the V twice but the radical curve once, so $\sqrt{x}$ has an inverse ($y = x^2$, $x \ge 0$) and $|x|$ does not.

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4. CLEP Exam Traps & Common Errors

• Trap 1: Writing $\sqrt{x^2} = x$. The correct simplification is $|x|$. Only when you already know $x \ge 0$ may you drop the bars.
• Trap 2: Getting the shift direction backwards. In $|x + 4|$ the vertex is at $x = -4$, not $x = 4$. Set the interior expression equal to zero and solve.
• Trap 3: Forgetting that a radical's range is bounded on one side. For $f(x) = a\sqrt{\cdots} + k$, the range always starts (or ends) at $k$ — it is never all real numbers.
• Trap 4: Treating $|x|$ as differentiable everywhere. The V has no single slope at its vertex. On a graph-matching item, the presence of a sharp corner rules out every polynomial choice.
• Trap 5: Ignoring the radicand when the radical sits in a denominator. For $f(x) = \dfrac{1}{\sqrt{x - 3}}$ the radicand must be strictly positive, so the domain is $(3, \infty)$, not $[3, \infty)$.