Lesson 10 of 11Approximate intersections without inventing solutions
Combine graph candidates, domains and controlled interval refinement across function families.
Try
A graph suggests that y=x³ and y=x+1 cross between x=1 and x=2. Rather than guessing a long decimal, subtract their heights. Define the residual h(x)=x³−x−1. An intersection occurs exactly when h(x)=0. At x=1 the residual is −1, and at x=2 it is 5.
The negative value says the cubic lies below the line; the positive value says it lies above. Because these functions are continuous throughout the interval, a zero lies between. This conclusion uses continuity, not just opposite signs at two arbitrary inputs.
Predict
Predict what happens at the midpoint 1.5: h(1.5)=3.375−1.5−1=0.875. The root therefore lies between 1 and 1.5. At 1.25 the residual is −0.296875, so retain [1.25,1.5]. Repeating this bisection procedure halves the bracket width at each step.
For this example h(1.32)=−0.020032 and h(1.33)=0.022637. Thus a root is in (1.32,1.33), and the midpoint 1.325 has absolute error below 0.005. That error claim follows from the bracket, not from the number of decimal places displayed by a calculator. A small residual alone need not imply a small input error on a nearly flat graph.
Experiment
Use the table as a set of specified function pairs. Draw or inspect each pair on shared axes, record its domain, and choose a plausible search interval. The functions are defined by their formulas, not merely by a finite set of samples.
| Family pair | Intersection equation | Essential check |
|---|
| Linear and linear | 2x+1=−x+4 | One shared height |
| Polynomial and linear | x³=x+1 | Continuous residual on all real inputs |
| Rational and linear | 1/(x−1)=x | Exclude x=1 before multiplying |
| Absolute value and constant | ∣x−2∣=3 | Check both sides of the vertex |
| Exponential and linear | 2ˣ=x+2 | Separate search intervals may contain different roots |
| Logarithmic and linear | ln x=2−x | Require x>0 |
| Radical and linear | √(x+1)=x/2 | Require x≥0 before squaring |
For the exponential pair, x=2 is exact, while a second root lies between −2 and −1: the residual is positive at −2 and negative at −1. Finding one crossing does not establish that it is the only one. For the absolute-value pair, the two branches give x=−1 and x=5.
Understand
A sign-change guarantee needs a continuous residual on the whole closed interval. The rational residual 1/(x−1)−x changes from −2.5 at x=0.5 to 0.5 at x=1.5, but it is undefined at x=1. This interval cannot be used as a bisection certificate. In fact its true roots are (1±√5)/2, approximately −0.618 and 1.618, both outside that interval. The vertical asymptote creates a false visual crossing if a plotting program connects points across the gap.
For the radical example, both sides of √(x+1)=x/2 require x≥0 at a solution. Squaring gives x²−4x−4=0, so the algebraic candidates are 2±2√2. The negative candidate violates the original sign requirement. Only 2+2√2 works. Squaring is not automatically reversible because positive and negative numbers can have the same square.
The logarithmic residual ln x+x−2 is defined and strictly increasing for positive x: increasing x raises both terms. Its values at 1 and 2 have opposite signs, so it has exactly one root in that interval and none elsewhere. This proves uniqueness by a general property, rather than by checking many plotted samples.
Apply
The six fresh practice cases use different constants from these worked examples. Include the domain, residual or branch equations, a justified interval or exact expression, and original-equation checks. For the absolute-value case, draw the V and line and label both crossings. For the polynomial bracket, the rubric asks only for uniqueness inside the bracket unless you provide an additional global argument.
Supplied exponential calculator values have stated precision. Their distances from the target are much larger than their rounding error, so their signs are reliable. When a displayed value is too close to zero to determine its sign, increase precision or use an exact bound; do not manufacture a bracket from uncertain rounding.
Continue
Graphing, tables and successive approximation are complementary methods. A graph locates behavior, a table makes intermediate values inspectable, and interval reasoning controls error. None permits ignoring a domain restriction.
Optional depth: a repeated root such as the zero of (x−1)² touches the horizontal axis without changing sign. Bisection based only on endpoint signs may miss it. A complete solution search must also consider tangencies, monotonicity, branch structure or algebraic factorization; a bounded numerical search is not a proof that no other solutions exist.
