Determine which curve is on top in each interval
A bar that changes color
The curve is \(f(x) = 6x - x^2\) and the line is \(g(x) = 2x + 3\). The orange bar joins them at the marked \(x\), and the shaded region between them runs from the bound \(x = -1\) to the bound \(x = 4\). Drag the bar from left to right.
At each \(x\), the higher of the two graphs is the top curve. Where \(f\) is below \(g\), the region and the bar are red, because \(f(x) - g(x)\) is negative. Where \(f\) is above \(g\), the region is teal and the bar is orange, because the difference is positive.
The bar changes color at \(x = 1\) and at \(x = 3\), where the two curves change places. These crossings cut the region into three intervals: from \(-1\) to 1, from 1 to 3, and from 3 to 4.
Inside one interval the bar keeps its color. One input there, called a test point, shows which curve is on top on the whole interval.
The order changes only at a crossing
The two crossings cut the region into three intervals, and the label in each interval names the higher curve there.
Drag the bar across the region. It is orange where \(f\) is higher and red where \(g\) is higher, and its color changes only at a crossing.
Curves without breaks cannot change places unless they meet, so the order is the same across one interval.
Notation
List the crossings between the bounds in order. They cut the interval from \(a\) to \(b\) into smaller intervals.
\[a < \cy{c_1} < \cy{c_2} < b\]Choose one test point \(t\) in each interval. Substitute it into both formulas.
\[f(t) \quad\text{and}\quad g(t)\]The curve with the larger value at \(t\) is the top curve on the whole interval. The other curve is the bottom curve.
\[f(t) > g(t) \;\Longrightarrow\; f \text{ is on top}\]Top minus bottom is the length of the bar, so it is positive on the whole interval. This is the difference you integrate for the area.
\[\cy{\text{top} - \text{bottom}} \ge 0\]
When it fails
The rule needs curves without breaks. Press the button to put a jump in \(f\). The curve drops below \(g\) without meeting it, so the equation \(f(x) = g(x)\) has no solution there.
The order still changes at the jump, where the orange edge moves from \(f\) to \(g\). Treat a break like a crossing and test one point on each side of it.
Example 1: A square root and a line
Which curve is on top, \(f(x) = \sqrt{x}\) or \(g(x) = \tfrac{x}{2}\), on each interval from \(x = 0\) to \(x = 9\)?
Find the crossings. Set the formulas equal and square both sides. Then multiply by 4 and factor.
\[\begin{aligned} \sqrt{x} &= \tfrac{x}{2} \\ x &= \tfrac{x^2}{4} \\ x(x - 4) &= 0 \end{aligned}\]Both solutions satisfy the original equation. The solution \(x = 0\) is the left bound, so the only crossing inside is \(x = 4\), which cuts the region into two intervals.
Test \(x = 1\) in the first interval. The curve is higher, so \(f\) is on top from 0 to 4.
\[\sqrt{1} = 1 > \tfrac12\]Test \(x = \tfrac{25}{4}\) in the second interval. The line is higher, so \(g\) is on top from 4 to 9.
\[\sqrt{\tfrac{25}{4}} = \tfrac52 < \tfrac{25}{8}\]
Example 2: An exponential and a line
Which curve is on top, \(f(x) = 2^x\) or \(g(x) = x + 1\), on each interval from \(x = -1\) to \(x = 3\)?
The equation \(2^x = x + 1\) cannot be solved by algebra. The picture shows that the curves meet at \(x = 0\) and \(x = 1\), so substitute to confirm.
\[\begin{aligned} 2^0 &= 1 = 0 + 1 \\ 2^1 &= 2 = 1 + 1 \end{aligned}\]A tempting wrong move: The curve \(f\) is higher at both bounds, so \(f\) is on top on the whole interval.
\[\begin{aligned} f(-1) = \tfrac12 &> 0 = g(-1) \\ f(3) = 8 &> 4 = g(3) \end{aligned}\]At \(x = \tfrac12\), \(f\) is \(\sqrt2 \approx 1.41\) and \(g\) is 1.5. The line is above the curve between the crossings, so the order is not the same on the whole interval.
The crossings cut the interval into three intervals: from \(-1\) to 0, from 0 to 1, and from 1 to 3. Test one point in each interval.
At \(x = -\tfrac12\), \(f\) is higher. At \(x = \tfrac12\), \(g\) is higher. At \(x = 2\), \(f\) is higher.
\[\begin{aligned} f &\text{ on } [-1, 0] \\ g &\text{ on } [0, 1] \\ f &\text{ on } [1, 3] \end{aligned}\]
Practice
Which curve is on top, \(f(x) = \dfrac{6}{x}\) or \(g(x) = x + 5\), on each interval from \(x = \tfrac12\) to \(x = 4\)?
Find the crossing inside the bounds first. Then test one point on each side of it.
Solution
Set the formulas equal. Multiply both sides by \(x\), which is positive here, to clear the fraction. Then factor.
\[\begin{aligned} \tfrac6x &= x + 5 \\ x^2 + 5x - 6 &= 0 \\ (x + 6)(x - 1) &= 0 \end{aligned}\]The crossing is \(x = 1\), because \(x = -6\) is outside the bounds. At \(x = \tfrac12\), \(f\) is 12 and \(g\) is 5.5, so \(f\) is on top from \(\tfrac12\) to 1.
\[\tfrac{6}{1/2} = 12 > 5.5\]At \(x = 2\), \(f\) is 3 and \(g\) is 7, so \(g\) is on top from 1 to 4.
\[\tfrac62 = 3 < 7 = 2 + 5\]
Questions
Questions 1 and 2 use \(f(x) = x^3\) and \(g(x) = 4x\) from \(x = -3\) to \(x = 3\).
At \(x = 1\), which curve is on top?
- The line \(g(x) = 4x\)
- The cubic \(f(x) = x^3\)
- Neither, the curves are equal
Answer
The line \(g(x) = 4x\)
On which intervals is the cubic \(f\) on top?
- \(-2 < x < 0\) and \(2 < x \le 3\)
- \(0 < x < 2\)
- The whole interval from \(-3\) to 3
- \(-3 \le x < -2\) and \(0 < x < 2\)
Answer
\(-2 < x < 0\) and \(2 < x \le 3\)
The curves \(f(x) = e^x\) and \(g(x) = e^{2 - x}\) cross at \(x = 1\). Which statement is correct on the interval from 0 to 3?
- \(g\) is on top from 0 to 1, and \(f\) is on top from 1 to 3
- \(f\) is on top from 0 to 1, and \(g\) is on top from 1 to 3
- \(f\) is on top from 0 to 3
- \(g\) is on top from 0 to 3
Answer
\(g\) is on top from 0 to 1, and \(f\) is on top from 1 to 3
Summary
In each interval between crossings, the curve with the larger value at one test point is the top curve.
This works because curves without breaks change places only where they meet, so the order stays the same inside each interval.
Use it before you write top minus bottom, and treat a break in a curve like a crossing.
This is one of four skills in topic 8.6 of Which & Why: AP Calculus.