Reading a Derivative Graph
A graph of \(f'\) displays the values of the derivative of a function \(f\). Its vertical coordinate is not the value of \(f\); it is the rate at which \(f\) is changing at the corresponding input.
This distinction makes the graph useful in context. Once you know what the input and output variables represent, you can read a point on the graph as a rate at a particular input and decide whether that rate is positive, negative, or zero.
What the Axes Represent
Before interpreting a point, identify both axes. The horizontal axis gives the input to \(f\). The vertical axis gives the value of \(f'\), the rate of change of \(f\) with respect to that input.
The units on the vertical axis follow from the variables: derivative units are the units of \(f\) divided by the units of the input. For example, if \(f(t)\) is a volume measured in liters and \(t\) is time measured in minutes, then \(f'(t)\) is measured in liters per minute.
Suppose a graph of \(f'\) has \(t\) in minutes on the horizontal axis and \(f'(t)\) in liters per minute on the vertical axis. If the graph passes through \((4,-3)\), then \(f'(4)=-3\) liters per minute. The first coordinate identifies when to read; the second gives the rate. The minus sign is part of that rate.
Reading the Sign of the Rate
The graph's position relative to the horizontal axis reveals the sign of the derivative. A point above that axis has a positive vertical coordinate; a point below it has a negative vertical coordinate; a point on it has vertical coordinate zero.
| Location on the graph of \(f'\) | Derivative value | Contextual interpretation |
|---|---|---|
| Above the horizontal axis | \(f'(a)>0\) | \(f\) is increasing at input \(a\) at a positive instantaneous rate. |
| Below the horizontal axis | \(f'(a)<0\) | \(f\) is decreasing at input \(a\) at a negative instantaneous rate. |
| On the horizontal axis | \(f'(a)=0\) | The instantaneous rate of change of \(f\) is zero at input \(a\). |
These descriptions concern the rate of change of \(f\), not whether \(f(a)\) itself is positive or negative. A function can have a positive value while decreasing, or a negative value while increasing. The graph of \(f'\) answers a question about change, rather than giving the original function's value.
A Short Justification
The derivative at an input \(a\) is defined, when the limit exists, by the limit of the average rates of change over intervals that shrink toward \(a\):
A graph of \(f'\) plots this derivative value against its input. Therefore, at horizontal coordinate \(a\), its vertical coordinate is \(f'(a)\) by the meaning of the graph. In a contextual problem, the derivative definition gives the instantaneous rate, so reading that coordinate gives the rate at the specified input. Its sign indicates whether that instantaneous rate is positive, negative, or zero.
This reasoning also explains why a zero derivative needs careful interpretation. It means the instantaneous rate is zero at that input. By itself, it does not establish that \(f\) has a maximum or minimum there, or that \(f\) remains constant around that input.
Worked Example: Water Volume Decreasing
Worked Example: Water Volume Decreasing
A graph of \(V'(t)\) shows the rate of change of water volume \(V\), in liters, in a tank. The horizontal axis is time \(t\), in minutes, and the vertical axis is \(V'(t)\), in liters per minute. The plotted point at \(t=4\) is \((4,-3)\). Interpret the derivative at that input.
Step 1: Read the input and derivative value. The horizontal coordinate is \(4\), so the graph gives the derivative at \(t=4\). The vertical coordinate is \(-3\), so \(V'(4)=-3\).
Step 2: Attach the units. Since \(V\) is measured in liters and \(t\) in minutes, \(V'(t)\) is measured in liters per minute. Thus, \(V'(4)=-3\) liters per minute.
Step 3: State the meaning of the sign. The negative rate means the volume is decreasing at that instant. The tank's water volume is decreasing at \(3\) liters per minute when \(t=4\) minutes.
Check: The graph's point at input \(4\) is below the horizontal axis at vertical coordinate \(-3\). The interpretation says decreasing, consistent with a negative derivative, and retains the graph's rate units of liters per minute.
Worked Example: A Quantity with Zero Rate
Worked Example: A Quantity with Zero Rate
Let \(P(t)\) be the number of plants in a greenhouse, with time \(t\) measured in weeks. A graph of \(P'(t)\), in plants per week, passes through \((6,0)\). What does the graph say about the rate at \(t=6\) weeks?
Step 1: Read the derivative value. At horizontal coordinate \(t=6\), the vertical coordinate is \(0\). Therefore, \(P'(6)=0\) plants per week.
Step 2: Interpret the zero rate. At \(t=6\) weeks, the instantaneous rate of change in the number of plants is zero. At that instant, the number of plants is neither increasing nor decreasing.
Step 3: Limit the conclusion to what is shown. The point tells us the rate at that instant. It does not show that the number of plants is zero, nor does it establish that the number stays unchanged throughout the weeks around \(t=6\).
Check: The plotted point lies on the horizontal axis, so its vertical coordinate is \(0\); the stated rate is therefore \(0\) plants per week. The units are plants divided by weeks, as required for \(P'(t)\).
Worked Example: Production at a Specified Amount
Worked Example: Production at a Specified Amount
Suppose \(C(q)\) is the operating cost, in dollars, of producing \(q\) items. A graph of \(C'(q)\) has vertical units of dollars per item. At \(q=20\), the graph has vertical coordinate \(1.5\). At \(q=30\), it has vertical coordinate \(0\). Interpret each rate separately.
At \(q=20\): The horizontal coordinate selects the input \(q=20\), and the vertical coordinate gives \(C'(20)=1.5\) dollars per item. The operating cost is increasing at an instantaneous rate of \(1.5\) dollars per additional item at that production level.
Check at \(q=20\): The graph's coordinate is \((20,1.5)\), which is above the horizontal axis. The interpretation is an increasing cost, and the units are dollars per item.
At \(q=30\): The vertical coordinate is zero, so \(C'(30)=0\) dollars per item. At a production level of \(30\) items, the instantaneous rate of change in operating cost is zero.
Check at \(q=30\): The point \((30,0)\) lies on the horizontal axis, so the derivative value is zero. The rate units remain dollars per item even though the numerical value is zero.
In both statements, \(q\) is the input and the graph's vertical coordinate is a rate. Neither coordinate is the operating cost \(C(q)\) itself.
A Common Mistake: Reading the Wrong Graph Value
A frequent error is to treat the vertical coordinate on a graph of \(f'\) as though it were \(f(a)\). Consider the water-volume graph above. A student might say, “At \(t=4\), the tank contains \(-3\) liters.” That is not what the graph says: \(-3\) is the vertical coordinate of the graph of \(V'\), so it is measured in liters per minute, not liters.
The correction is to name the plotted quantity first. Since the graph is of \(V'\), its coordinate gives \(V'(4)=-3\) liters per minute. It says the volume is decreasing at that instant. To know the amount of water \(V(4)\), one would need information from a graph or formula for \(V\), not merely this point on the graph of \(V'\).
A Reliable Reading Routine
For each requested input, use the same sequence. It helps keep the graph reading, sign, units, and contextual statement aligned.
Confirm that the plotted quantity is \(f'\), not \(f\).
Find the specified horizontal coordinate \(a\).
This is \(f'(a)\), including its sign.
State function-units per input-unit, as indicated by the variables.
Describe the quantity as increasing, decreasing, or having zero instantaneous rate.
Why This Reading Matters
In an applied setting, the sign of a rate can answer a practical question directly. A positive derivative indicates that the measured quantity is increasing at the specified input; a negative derivative indicates that it is decreasing; and a zero derivative indicates no instantaneous change at that point. The units state how quickly the change is measured relative to the input.
The graph of \(f'\) therefore connects a visual location to a contextual statement. To read it accurately, keep three pieces of information together: the input, the vertical coordinate, and the units. Then interpret only what that derivative value warrants. In particular, a zero rate at one input is a statement about that instant, not a claim that the function's value is zero or remains fixed nearby.
Check Your Understanding
For each question, imagine that the stated values are read from a graph of the derivative. Include the contextual meaning and units where they are given.
- A graph of \(M'(t)\) is measured in kilograms per hour. At \(t=5\), its vertical coordinate is \(2\). What does this tell you about \(M\) at that input?
- A graph of \(H'(x)\) lies below the horizontal axis at \(x=8\). What is the sign of \(H'(8)\), and what does that sign mean for \(H\) at \(x=8\)?
- A graph of \(R'(t)\), measured in dollars per day, passes through \((3,0)\). State what is known about the instantaneous rate and one conclusion that cannot be drawn from this point alone.
- A student says that a point on the graph of \(A'\) with vertical coordinate \(-4\) means \(A\) has value \(-4\). What has the student confused?
- If the input is measured in minutes and \(f\) is measured in liters, what are the units of the vertical coordinate on a graph of \(f'\)?
