What Does It Mean for Two Graphs to Match?
A graph of a function \(f\) shows the function's values: at input \(x=a\), its height is \(f(a)\). A graph of the derivative \(f'\) shows a different quantity. At the same input \(a\), its height is \(f'(a)\), the slope of the tangent line to the graph of \(f\) there, when that derivative exists.
Matching graphs therefore means checking a specific relationship: the height of the proposed derivative graph at an input must equal the tangent slope of the proposed function graph at that input. The function's height itself is not the quantity to compare with the derivative graph.
In a graph-matching problem, the candidates may be drawn without formulas. You can still test them by reading marked inputs, finding tangent slopes on the function graph, and comparing those slopes with the derivative-graph heights at exactly the same inputs.
Why Tangent Slopes Give the Test
The derivative at \(a\) is defined by the limit of secant slopes, if that limit exists. For nonzero \(h\), the secant line joins the points \((a,f(a))\) and \((a+h,f(a+h))\), so its slope is \(\frac{f(a+h)-f(a)}{h}\). As \(h\) approaches zero, the second point approaches the first. When the limit exists, it is the tangent slope and is the derivative value.
This gives the matching rule directly: the proposed derivative graph must have height \(f'(a)\) above the input \(a\). A graph that has the right height at a different input does not pass the test. Nor does a graph pass because its height happens to equal \(f(a)\); function values and tangent slopes are different quantities.
A Reliable Matching Procedure
Use the same horizontal coordinate on the candidate function graph and candidate derivative graph.
If a tangent line is drawn, use two points on that line and calculate rise divided by run. If its slope is labeled, record that value.
At that same input, determine the proposed derivative value.
A disagreement rules out the proposed match. Check several marked inputs before deciding which candidates remain consistent.
For a drawn tangent, the slope is computed as \(\frac{\text{change in vertical coordinate}}{\text{change in horizontal coordinate}}\). The order of the two points must be consistent in numerator and denominator. A positive slope means the tangent rises from left to right; a negative slope means it falls. A horizontal tangent has slope zero.
Worked Example: Matching Two Curves to Two Derivative Lines
Worked Example: Matching Two Curves to Two Derivative Lines
Suppose a diagram gives two candidate function curves, \(F\) and \(G\), and two candidate derivative lines, \(P\) and \(Q\). The marked inputs are \(-1\), \(1\), and \(2\). The tangent lines on \(F\) pass through the following pairs of points:
- At \(x=-1\): \((-1,1)\) and \((0,-1)\).
- At \(x=1\): \((1,1)\) and \((2,3)\).
- At \(x=2\): \((2,4)\) and \((3,8)\).
Calculate the three tangent slopes for \(F\). At \(x=-1\), the change in height is \(-1-1=-2\), and the change in input is \(0-(-1)=1\), so the slope is \(\frac{-2}{1}=-2\). At \(x=1\), the changes are \(3-1=2\) and \(2-1=1\), so the slope is \(\frac{2}{1}=2\). At \(x=2\), the changes are \(8-4=4\) and \(3-2=1\), so the slope is \(\frac{4}{1}=4\).
The derivative-line candidates are \(P(x)=2x\) and \(Q(x)=-2x\). Evaluate each at all three marked inputs:
The three values from \(P\) are \(-2,2,4\), exactly the slopes read from \(F\). Thus \(P\) passes all three checks for \(F\). The values from \(Q\) are \(2,-2,-4\), which disagree with those slopes, so \(Q\) does not match \(F\) at the marked inputs.
For the curve \(G\), suppose the tangent lines at \(x=-1,1,2\) have slopes \(2,-2,-4\), respectively. These are precisely the values of \(Q\) at those inputs, while \(P\) has the opposite values. Therefore \(Q\) passes the marked-input checks for \(G\), and \(P\) fails them.
Check: The matches give the required value at every tested input: \(P(-1)=-2\), \(P(1)=2\), and \(P(2)=4\) agree with \(F\)'s slopes; \(Q(-1)=2\), \(Q(1)=-2\), and \(Q(2)=-4\) agree with \(G\)'s slopes. The rejected pairings disagree at each of these inputs.
Worked Example: Different Heights, the Same Derivative
Worked Example: Different Heights, the Same Derivative
Two candidate function curves, \(U\) and \(V\), have tangent slopes \(0\), \(2\), and \(4\) at \(x=0\), \(x=1\), and \(x=2\), respectively. The curves are vertically separated in the diagram: their heights are not the same. A proposed derivative graph is \(R(x)=2x\). Does it pass the tests for both curves?
At \(x=0\), the tangent slope for each curve is \(0\), and \(R(0)=2(0)=0\). At \(x=1\), each tangent slope is \(2\), and \(R(1)=2(1)=2\). At \(x=2\), each tangent slope is \(4\), and \(R(2)=2(2)=4\).
Thus the proposed derivative graph agrees with both curves at all three marked inputs, even though the function curves have different heights. For example, if \(U(1)=1\) and \(V(1)=6\), those heights do not change the comparison: the required quantity at \(x=1\) is the tangent slope, which is \(2\), not either function value.
Check: The three derivative values are \(R(0)=0\), \(R(1)=2\), and \(R(2)=4\), matching the stated tangent slopes for each curve. This example illustrates why matching depends on slopes, not vertical position.
Worked Example: A Corner Prevents a Match
Worked Example: A Corner Prevents a Match
Suppose a candidate function graph has a sharp corner at \(x=0\). On the left of the corner, its graph is the line \(y=-x\); on the right, its graph is the line \(y=x\). A candidate derivative graph is drawn with the value \(S(0)=0\). Can this be the derivative graph of the function at the marked input?
From the left, the secant slope from \((0,0)\) to \((-h,h)\), for \(h>0\), is \(\frac{h-0}{-h-0}=-1\). From the right, the secant slope from \((0,0)\) to \((h,h)\) is \(\frac{h-0}{h-0}=1\). The two one-sided slopes are different, so there is no single tangent slope at the corner and the ordinary derivative at \(0\) does not exist.
Consequently, the proposed value \(S(0)=0\) cannot be the derivative value there. The issue is not that zero is the wrong slope to estimate from the graph; rather, no derivative value exists at that input.
Check: The left-hand slopes equal \(-1\), and the right-hand slopes equal \(1\). Since \(-1\ne 1\), the two-sided derivative does not exist, so a derivative graph cannot assign the function a derivative value at \(0\).
What Marked-Input Checks Establish
Each marked input provides a necessary test for a proposed match. If the tangent slope is \(3\) at \(a\), but the candidate derivative graph has height \(2\) at \(a\), that pairing is impossible. One disagreement is enough to reject it.
Passing several checks is useful, but it does not by itself prove that two entire graphs match for every input. The checks only establish agreement at the inputs that were examined. Two candidate derivative graphs could have the same values at those marked inputs and differ elsewhere. A global conclusion requires enough additional information to establish agreement throughout the domain in question.
There is a related limitation in reading drawn graphs: a slope estimated from a sketch may be approximate. If the problem supplies a grid, labeled tangent, or exact coordinates, use them to calculate the slope rather than relying on visual steepness alone. When only a sketch is available, report the comparison at the level of precision the sketch supports.
A Common Mistake: Comparing Heights Instead of Slopes
Suppose a function graph passes through \((2,5)\), and the tangent there rises \(3\) units for every \(1\) unit of horizontal movement. The function value is \(f(2)=5\), while the tangent slope is \(3\). A common incorrect calculation is to say that the derivative graph must have value \(5\) at \(x=2\), because the function graph has height \(5\) there. That compares the wrong quantities.
The correction is to use the tangent line: its slope is \(\frac{3}{1}=3\), so the derivative graph must have height \(f'(2)=3\) at input \(2\). The function height \(5\) records the point on the curve; it does not give the tangent slope.
Check Your Understanding
Use the tangent-slope test to answer each question.
- At \(x=3\), a tangent line passes through \((3,2)\) and \((5,8)\). What value must a matching derivative graph have at \(x=3\)?
- A candidate derivative graph has value \(-2\) at \(x=1\), while the function's tangent slope there is \(4\). What conclusion follows about that pairing?
- Two function curves have different heights at \(x=0\), but both have tangent slope \(1\) there. What derivative value should a matching graph have at \(x=0\)?
- A function has different left-hand and right-hand slopes at a marked input. Can its ordinary derivative graph assign a derivative value at that input? Explain.
- If a candidate derivative graph agrees with a function's tangent slopes at three marked inputs, what has been established, and what has not necessarily been established?
