Two Different Questions
A function value and a derivative value answer different questions. The function value \(f(a)\) tells you the output at the input \(a\). The derivative \(f'(a)\), when it exists, tells you the instantaneous rate of change at that input: whether the function is increasing or decreasing there, and how steeply.
Confusing these quantities can lead to a basic but consequential mistake: deciding that \(f(a)\) must be positive because \(f'(a)\) is positive. That conclusion does not follow. A function can be below zero while increasing, or above zero while decreasing.
Reading the Signs on a Graph
On a graph of \(y=f(x)\), the sign of \(f(a)\) is read from the point’s vertical position. If the point \((a,f(a))\) is above the horizontal axis, then \(f(a)>0\); if it is below the axis, then \(f(a)<0\). A point on the axis has \(f(a)=0\).
The sign of \(f'(a)\) is read from the slope of the tangent line at that point, assuming the derivative exists. A tangent that rises from left to right has positive slope, so \(f'(a)>0\). A tangent that falls from left to right has negative slope, so \(f'(a)<0\). A horizontal tangent has derivative zero.
These are independent visual features: the point’s height and the tangent’s tilt. A point can sit below the axis while its tangent rises, just as a person walking uphill can still be below sea level.
| Function value | Derivative | Interpretation |
|---|---|---|
| \(f(a)>0\) | \(f'(a)>0\) | Above zero and increasing locally |
| \(f(a)<0\) | \(f'(a)>0\) | Below zero and increasing locally |
| \(f(a)>0\) | \(f'(a)<0\) | Above zero and decreasing locally |
| \(f(a)<0\) | \(f'(a)<0\) | Below zero and decreasing locally |
Worked Example: Four Sign Combinations
Worked Example: Four Sign Combinations
Consider the four linear functions \(f_1(x)=5+2x\), \(f_2(x)=-5+2x\), \(f_3(x)=5-2x\), and \(f_4(x)=-5-2x\). Classify the function value and derivative at \(x=0\).
For \(f_1\), the value is \(f_1(0)=5+2(0)=5>0\). Its slope is \(2\), so \(f_1'(0)=2>0\). This function is above zero and increasing at \(0\).
For \(f_2\), \(f_2(0)=-5+2(0)=-5<0\), while its slope is \(2\), so \(f_2'(0)=2>0\). It is below zero and increasing.
For \(f_3\), \(f_3(0)=5-2(0)=5>0\), while its slope is \(-2\), so \(f_3'(0)=-2<0\). It is above zero and decreasing.
For \(f_4\), \(f_4(0)=-5-2(0)=-5<0\), and its slope is \(-2\), so \(f_4'(0)=-2<0\). It is below zero and decreasing.
Check: Each function is a line, so its derivative equals its constant slope. Substituting \(x=0\) into each formula gives the values \(5,-5,5,-5\), respectively. All four possible sign pairings occur, confirming that the two signs are not interchangeable.
What a Derivative Sign Does Tell You
Although the derivative’s sign does not tell you the function’s sign, it does give a precise local comparison with the value at the point. Suppose \(f\) is differentiable at an interior point \(a\). If \(f'(a)>0\), then for inputs sufficiently close to \(a\), values just to the right are greater than \(f(a)\), while values just to the left are less than \(f(a)\). If \(f'(a)<0\), those comparisons reverse.
This result compares nearby function values with \(f(a)\); it does not compare \(f(a)\) with zero. Here is a proof for the positive-derivative case. By the definition of derivative, the difference quotient approaches \(f'(a)\) as \(h\) approaches zero. Since \(f'(a)>0\), choose the tolerance \(f'(a)/2\). For all sufficiently small nonzero \(h\), the quotient is within that tolerance of \(f'(a)\), so it is positive:
When \(h>0\), multiplying the positive quotient by \(h\) gives \(f(a+h)-f(a)>0\). When \(h<0\), multiplying by \(h\) reverses the inequality, giving \(f(a+h)-f(a)<0\). This proves the claim for \(f'(a)>0\). If \(f'(a)<0\), apply the same reasoning with a tolerance of \(|f'(a)|/2\): the quotient is negative for sufficiently small nonzero \(h\), and multiplication by positive or negative \(h\) gives the reversed comparisons.
The condition that \(a\) is an interior point matters because the argument uses inputs on both sides of \(a\). The conclusion is also local: it concerns inputs close enough to \(a\), not every input in the function’s domain. If \(f'(a)=0\), this argument gives no direction of change.
Worked Example: A Positive Population That Is Decreasing
Worked Example: A Positive Population That Is Decreasing
Suppose a model for a population is \(P(t)=1200-30t\), where \(t\) is measured in hours and \(P(t)\) is measured in individuals. Consider \(0\leq t\leq20\), and find the population and its rate of change at \(t=10\).
Substitution gives \(P(10)=1200-30(10)=1200-300=900\) individuals, so \(P(10)>0\). The graph is a line with slope \(-30\), so \(P'(10)=-30\) individuals per hour. Thus the population is positive at that time but decreasing.
The model remains positive throughout the stated interval: it is decreasing, and its smallest value on the interval is \(P(20)=1200-30(20)=600\) individuals. Check: Over the next hour, \(P(11)=1200-30(11)=870\), a change of \(870-900=-30\) individuals. This agrees with the rate \(-30\) individuals per hour. The negative derivative describes the decrease; it does not mean the population itself is negative.
Worked Example: Rising While Below a Reference Level
Worked Example: Rising While Below a Reference Level
Let \(E(t)=-12+1.5t\) represent elevation in metres relative to a chosen reference level, with \(t\) in minutes. At \(t=4\), calculate the elevation and rate of change, and interpret their signs.
The elevation is \(E(4)=-12+1.5(4)=-12+6=-6\) metres. The line has slope \(1.5\) metres per minute, so \(E'(4)=1.5>0\). The location is six metres below the reference level but is rising.
To check both the rate and the interpretation, at \(t=5\) the elevation is \(E(5)=-12+1.5(5)=-12+7.5=-4.5\) metres. The change is \(-4.5-(-6)=1.5\) metres in one minute, matching the stated rate. Solving \(E(t)=0\) gives \(1.5t=12\), hence \(t=8\) minutes; the elevation is still negative at \(t=4\), even though it is increasing toward zero.
Common Mistake: Treating “Increasing” as “Positive”
Consider \(f(x)=-4+3x\) at \(x=0\). A mistaken interpretation says, “The derivative is positive, so the function is positive.” The calculation disproves it: \(f(0)=-4+3(0)=-4<0\), while the line’s slope is \(f'(0)=3>0\). The function is increasing from a negative value.
The zero of this line is found from \(-4+3x=0\), so \(x=4/3\). At \(x=0\), the graph is below the horizontal axis; its positive slope means it rises as \(x\) increases. It does not become positive until after it crosses the axis. The correction is to ask two separate questions: “What is the value?” and “Which way is the function changing?”
Why the Distinction Matters
In applications, zero often has a meaningful reference: zero elevation, zero profit, zero temperature change, or zero position relative to an origin. The derivative describes how the modeled quantity changes, with its own units. For example, a population measured in individuals can be positive while its rate, measured in individuals per hour, is negative. These signs answer different questions and should not be substituted for one another.
A reliable interpretation keeps three pieces of information distinct: the value \(f(a)\), its sign relative to zero, and the derivative \(f'(a)\), whose sign indicates local direction of change. On a graph, read height and tangent slope separately. In a context, state what the zero reference means and include the units of the rate.
Check Your Understanding
For each question, distinguish the function’s level from its direction of change.
- If \(f(a)<0\) and \(f'(a)>0\), what can you say about the value at \(a\) and its local direction of change?
- On a graph, what feature tells you the sign of \(f(a)\), and what feature tells you the sign of \(f'(a)\)?
- A population is \(500\) individuals and has rate of change \(-12\) individuals per day. Is the population negative or decreasing, or both?
- What local comparison follows when \(f'(a)<0\), and what assumption is needed about \(a\)?
- Why does \(f'(a)=0\) alone not determine whether \(f(a)\) is positive, negative, or zero?
