4.5 Graphical Solutions: Reading f(x) = g(x) and f(x) < g(x)

4.5 Graphical Solutions: Reading f(x) = g(x) and f(x) < g(x)

[!NOTE] Regents Standard Alignment: AII-A.REI.11 - explain why the $x$-coordinates of the points where the graphs of $y = f(x)$ and $y = g(x)$ intersect are the solutions of the equation $f(x) = g(x)$. NYSED's Algebra II note extends the standard twice: "Students are recognizing the graphical solution to $f(x) = g(x)$ and interpreting the solution in context. Additional functions include square root, cube root and trigonometric functions. Students also need to be able to recognize the graphical solution to $f(x) < g(x)$ or $f(x) \leq g(x)$."

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1. Why Intersections Are Solutions

Take any point $(a, b)$ that lies on both graphs. Lying on $y = f(x)$ means $b = f(a)$. Lying on $y = g(x)$ means $b = g(a)$. Two things equal to the same quantity are equal to each other, so

$f(a) = b = g(a) \quad \Longrightarrow \quad f(a) = g(a).$

That is precisely the statement that $x = a$ solves $f(x) = g(x)$. The logic runs backwards just as cleanly: if $x = a$ solves the equation, then the point $(a, f(a))$ satisfies both equations and is therefore an intersection point.

Two consequences follow immediately, and both are scored.

1. The solution is the $x$-coordinate, not the point. If the curves meet at $(3, 7)$, the solution of $f(x) = g(x)$ is $x = 3$. Writing $(3, 7)$ answers a system question, not an equation question. Read the prompt: "solve the equation" wants inputs; "solve the system" wants ordered pairs.
2. No intersection means no real solution. A pair of curves that never meet corresponds to an equation with an empty real solution set.

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2. From Equation to Inequality

Once you can read equality off a graph, inequality is one more observation: compare vertical position.

The boundaries of these intervals are always intersection points or the edges of a domain. Missing a domain edge is the single most common error on this item type.

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3. Worked Example 1: A Quadratic and a Line

Problem: Let $f(x) = x^2 - 4$ and $g(x) = 3x$. Determine the solutions of $f(x) = g(x)$ and the solution set of $f(x) < g(x)$.

• Intersections. $x^2 - 4 = 3x \Rightarrow x^2 - 3x - 4 = 0 \Rightarrow (x - 4)(x + 1) = 0$, so $x = -1$ and $x = 4$. The curves meet at $(-1, -3)$ and $(4, 12)$.
• Where is the parabola below the line? Both functions are defined for every real number, so the only boundaries are the two intersections. Test one point inside and one point outside:
  • At $x = 0$: $f(0) = -4$ and $g(0) = 0$, so $f < g$. Inside.
  • At $x = 5$: $f(5) = 21$ and $g(5) = 15$, so $f > g$. Outside.
• Answer. $f(x) = g(x)$ at $x = -1$ and $x = 4$; $f(x) < g(x)$ on the open interval $(-1, 4)$.

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4. Worked Example 2: A Radical and a Line - Why Graphing Never Lies

Problem: Solve $\sqrt{x + 4} = x - 2$ graphically, and compare with the algebraic solve.

• Graphically. $y = \sqrt{x + 4}$ starts at $(-4, 0)$ and rises slowly; $y = x - 2$ is a line crossing the axis at $(2, 0)$. They meet once, at $(5, 3)$. So the graphical solution is $x = 5$.
• Algebraically. Squaring gives $x + 4 = x^2 - 4x + 4$, then $x^2 - 5x = 0$, then $x(x - 5) = 0$, producing candidates $x = 0$ and $x = 5$.
• Reconciling the two. The candidate $x = 0$ is extraneous: $\sqrt{0 + 4} = 2$ while $0 - 2 = -2$. On the graph, $x = 0$ is where the radical sits at height $2$ and the line sits at height $-2$ - the two curves are nowhere near each other there. The squaring step created a solution of the companion equation $\sqrt{x+4} = -(x-2)$, and the graph simply never had it.

[!IMPORTANT] A graphical solution cannot produce an extraneous root. An intersection exists only if both functions actually take the same value at that input. That makes graphing the fastest possible independent check after an algebraic solve - which is exactly how NYSED expects the graphing calculator to be used on an "algebraically determine" prompt.

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5. Worked Example 3: A Logarithm and an Absolute Value - Domain Sets an Endpoint

Problem: Given $f(x) = -|x|$ and $g(x) = \log_2(x)$, determine the interval that best represents the solution set of $g(x) > f(x)$.

• Domain first. $\log_2(x)$ requires $x > 0$. Nothing to the left of the origin can possibly belong to the solution set, regardless of what the graphs look like there.
• Behavior near zero. As $x \to 0^+$, $\log_2(x) \to -\infty$ while $-|x| \to 0$. So immediately to the right of zero the logarithm is below the absolute value.
• Behavior for large $x$. At $x = 1$, $\log_2(1) = 0$ and $-|1| = -1$, so the logarithm is above. Somewhere between $0$ and $1$ the curves cross.
• Locate the crossing. Using 2nd TRACE -> 5: intersect on $Y_1 = \log(x)/\log(2)$ and $Y_2 = -\mathrm{abs}(x)$ gives $x \approx 0.64$.
• Answer. $g(x) > f(x)$ on $(0.64, \infty)$ - open at $0.64$ because the inequality is strict, and bounded below by the crossing rather than by the domain edge, since the logarithm sits below the absolute value on $(0, 0.64)$.

Notice how the wrong answer $(-0.64, \infty)$ is built: it comes from solving without ever asking where $\log_2(x)$ is defined.

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6. Worked Example 4: A Trigonometric Reading

Problem: Determine all solutions of $\cos(x) = -\tfrac{1}{2}$ on $[0, 2\pi)$ by reading the graph of $y = \cos(x)$ against the horizontal line $y = -\tfrac{1}{2}$.

• The cosine curve starts at $(0, 1)$, falls through the midline at $\frac{\pi}{2}$, reaches $-1$ at $\pi$, and returns to $1$ at $2\pi$.
• The horizontal line $y = -\frac{1}{2}$ therefore cuts the curve twice on $[0, 2\pi)$: once while cosine is decreasing, once while it is increasing.
• The reference angle satisfying $\cos\theta_R = \frac{1}{2}$ is $\frac{\pi}{3}$. Cosine is negative in Quadrants II and III, so the two crossings are at

$x = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \quad \text{and} \quad x = \pi + \frac{\pi}{3} = \frac{4\pi}{3}.$

• Extending to an inequality. $\cos(x) < -\frac{1}{2}$ is where the curve dips below the line, which on $[0, 2\pi)$ is the single interval $\left(\frac{2\pi}{3}, \frac{4\pi}{3}\right)$.

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7. Calculator Workflow and Reporting Checklist

Y1 = left side          Y2 = right side
GRAPH  ->  adjust WINDOW until every crossing is visible
2nd TRACE -> 5: intersect  ->  First curve? Second curve? Guess?
Repeat once per visible crossing

• Set a window you can defend. A crossing outside the viewing rectangle is a missed solution, and the calculator will not warn you.
• Report inputs. Intervals of $x$, not of $y$.
• Check both domains at every endpoint. Even radicals need a non-negative radicand; logarithms need a positive argument; rational expressions need nonzero denominators.
• Match the bracket to the symbol. Strict inequalities get parentheses; $\leq$ and $\geq$ get square brackets at intersection points, but never at a point excluded by a domain.
• Interpret in context when asked. NYSED's note requires interpreting the solution in context, so if $x$ is time in hours, finish with a sentence naming the hours.