Wemazi

Topic 8.6

Find every relevant crossing

Builds on calculating area with horizontal strips. 6 minutes.

Where the bar shrinks to a point

The curve is \(f(x) = x^3 - 3x^2\) and the line is \(g(x) = x - 3\). The teal region between them runs between the bounds \(x = 0\) and \(x = 3.5\). The orange bar joins the two curves. Drag the bar to the right.

At \(x = 1\) and at \(x = 3\) the bar shrinks to a point, because the curves have the same height there. These inputs are crossings, and a dot marks each one.

The curves also cross at \(x = -1\), left of the region. That dot is faint because the crossing is outside the bounds, so it does not cut the region.

The crossings that matter are the ones between the two bounds.

Every crossing, inside and outside

The two orange lines are the bounds \(a\) and \(b\). Each dot marks a crossing, an input where the two curves meet.

Drag the sliders to move the bounds. When a bound passes a dot, the count changes. A dot outside the bounds is faint, because that crossing does not cut the region.

If a bound lies on a crossing, the region ends there, so that crossing does not cut the region either.

Notation

  1. A crossing is an input \(x\) where the two curves have the same height, so the orange bar between them has length 0.

    \[f(x) = g(x)\]

  2. To find the crossings, move every term to one side and solve. The difference \(f(x) - g(x)\) is the length of the bar, so each solution is an input where the bar has length 0.

    \[\cy{f(x) - g(x)} = 0\]

    Do not divide by a factor that can be 0, because the division loses a solution.

  3. Keep the solutions between the bounds \(a\) and \(b\), because a solution outside them does not touch the region.

    \[a < x < b\]

  4. Substitute each solution into \(f(x) = g(x)\). Both sides must have the same value, because solving can produce an input that is not a crossing.

    Count the crossings before you write any integral.

When it fails

To solve \(\sqrt{2x + 3} = x\), you square both sides. This gives \(x^2 = 2x + 3\), which has the solutions \(x = 3\) and \(x = -1\).

Squaring loses the sign of \(\sqrt{2x + 3}\), which is never negative, so the squared equation also finds where the line meets the lower half, \(y = -\sqrt{2x + 3}\). Press the button to see that half meet the line at \((-1, -1)\), labeled in red. The curve itself is 1 at \(x = -1\), not \(-1\), so only \(x = 3\) is a crossing.

Example 1: A crossing outside the bounds

Find the crossings of \(f(x) = 3 - x^2\) and \(g(x) = 2x\) for the region from \(x = 0\) to \(x = 4\).

  1. Set the two formulas equal.

    \[3 - x^2 = 2x\]

  2. Move every term to one side and factor.

    \[\begin{aligned} x^2 + 2x - 3 &= 0 \\ (x + 3)(x - 1) &= 0 \end{aligned}\]

  3. The solutions are \(x = -3\) and \(x = 1\). The bounds are 0 and 4, so \(x = -3\) is outside the region and its dot is faint. Only \(x = 1\) cuts the region.

  4. Substitute \(x = 1\) into both formulas. Both give 2, so the curves meet at \((1, 2)\).

    \[3 - 1^2 = 2, \qquad 2(1) = 2\]

Example 2: Three crossings of two sine curves

Find the crossings of \(f(x) = \sin x\) and \(g(x) = \sin 2x\) for the region from \(x = -\pi\) to \(x = \pi\).

  1. Set the formulas equal. The double-angle identity writes \(\sin 2x\) as \(2\sin x\cos x\).

    \[\sin x = 2\sin x\cos x\]

  2. A tempting wrong move: Divide both sides by \(\sin x\).

    \[1 = 2\cos x \;\Longrightarrow\; x = \pm\tfrac{\pi}{3}\]

    The curves also meet at the origin, where both are 0. Dividing by \(\sin x\) assumes it is not 0, so this crossing is lost.

  3. Move every term to one side and factor instead. Then set each factor equal to 0.

    \[\begin{aligned} 2\sin x\cos x - \sin x &= 0 \\ \sin x\,(2\cos x - 1) &= 0 \end{aligned}\]

  4. The factor \(\sin x\) is 0 at \(x = -\pi\), 0 and \(\pi\). The factor \(2\cos x - 1\) is 0 where \(\cos x = \tfrac12\), at \(x = \pm\tfrac{\pi}{3}\).

  5. The bounds \(-\pi\) and \(\pi\) are crossings, but they are the ends of the region, so they do not cut it. The crossings inside are \(-\tfrac{\pi}{3}\), 0 and \(\tfrac{\pi}{3}\).

    \[\cy{x = -\tfrac{\pi}{3},\ 0,\ \tfrac{\pi}{3}}\]

Practice

Find the crossings of \(f(x) = x^4\) and \(g(x) = 5x^2 - 4\) for the region from \(x = -3\) to \(x = 3\).

Drag the bar to count where it shrinks to a point. Then move every term to one side. The result has only even powers, so treat \(x^2\) as one unknown and factor.

Solution

  1. Move every term to one side. The expression is a quadratic in \(x^2\), so factor it.

    \[\begin{aligned} x^4 - 5x^2 + 4 &= 0 \\ (x^2 - 1)(x^2 - 4) &= 0 \end{aligned}\]

  2. Each factor gives two solutions: \(x^2 = 1\) gives \(x = \pm 1\), and \(x^2 = 4\) gives \(x = \pm 2\).

  3. All four solutions lie between \(-3\) and 3, so all four cut the region. At \(x = 2\), both formulas give 16, which confirms that solution.

    \[2^4 = 16, \qquad 5(2)^2 - 4 = 16\]

Questions

Questions 1 and 2 use the curves \(f(x) = x^2\) and \(g(x) = 6 - |x|\).

  1. The bounds are \(x = -3\) and \(x = 3\). How many crossings lie between them?

    • 2
    • 1
    • 3
    • 4
    Answer

    2

  2. The bounds are \(x = 0\) and \(x = 3\). At which input is there a crossing between them?

    • \(x = 2\)
    • \(x = -2\)
    • \(x = 4\)
    • \(x = 0\)
    Answer

    \(x = 2\)

  3. Where do \(y = \dfrac{8}{x}\) and \(y = 6 - x\) cross for \(x\) between 3 and 5?

    • \(x = 4\)
    • \(x = 2\) and \(x = 4\)
    • \(x = 2\)
    • \(x = 3\) and \(x = 5\)
    Answer

    \(x = 4\)

Summary

You set the two formulas equal, move every term to one side, factor, and keep the solutions between the bounds.

This works because the top curve can change only where the curves meet, and a crossing outside the bounds does not touch the region.

Use it before you choose the top curve or write any integral.

This is one of four skills in topic 8.6 of Which & Why: AP Calculus.