Reading a Function’s Trend from Its Derivative Graph
A graph of \(f'\) is not a graph of \(f\): its height at each input gives the slope of \(f\) there. But that slope information lets us describe how \(f\) changes. When \(f'\) stays positive, \(f\) rises; when \(f'\) stays negative, \(f\) falls. As the values of \(f'\) change, the local slopes of \(f\) change too.
There are two separate questions to ask while reading a derivative graph. First, is \(f'\) above or below the horizontal axis? This tells us whether \(f\) rises or falls. Second, are the values of \(f'\) becoming larger or smaller in magnitude? This tells us whether the slopes of \(f\) are becoming steeper or flatter. Keeping these questions separate prevents a common misreading.
Direction of Change: Above or Below Zero
The horizontal axis on a derivative graph represents \(f'(x)=0\). Where the derivative graph lies above that axis, \(f'(x)>0\); where it lies below, \(f'(x)<0\). On an interval, these signs tell us whether \(f\) is increasing or decreasing, provided the usual conditions of the Mean Value Theorem hold.
“Nondecreasing” means that whenever \(x_1<x_2\), we have \(f(x_1)\leq f(x_2)\); “nonincreasing” reverses that inequality. The weak versions of the sign test allow the derivative to be zero. A derivative value of zero at one point means the tangent there is horizontal, but does not by itself say that \(f\) stops changing over a surrounding interval.
Why the Sign Test Works
The Mean Value Theorem says that if \(f\) is continuous on \([x_1,x_2]\) and differentiable on \((x_1,x_2)\), then there is some \(c\in(x_1,x_2)\) such that the average rate of change across the interval equals the derivative at \(c\):
For \(x_1<x_2\), the denominator is positive. So if \(f'\) is positive throughout the interval, the quotient is positive, which forces \(f(x_2)>f(x_1)\). If \(f'\) is negative throughout, the quotient is negative, so \(f(x_2)<f(x_1)\). If \(f'\) is nonnegative, the quotient cannot be negative, so \(f(x_2)\geq f(x_1)\). This proves the sign test under its stated conditions.
The theorem turns information about instantaneous slopes into a conclusion about function values across an interval. In practice, when the derivative graph is given, divide the input range wherever \(f'\) crosses or touches zero, then determine the sign on each resulting interval.
Steeper or Flatter: Track the Magnitude
The height of the derivative graph is the signed slope, not steepness by itself. Steepness is measured by \(\lvert f'(x)\rvert\). Thus, a function can be falling while becoming less steep, or falling while becoming more steep. The sign of \(f'\) answers whether \(f\) rises or falls; changes in \(\lvert f'\rvert\) answer whether the tangent slopes become steeper or flatter.
| Derivative values | Trend of \(f\) | Trend in steepness |
|---|---|---|
| Positive and increasing | Rising | Steeper |
| Positive and decreasing | Rising | Flatter |
| Negative and increasing toward zero | Falling | Flatter |
| Negative and decreasing away from zero | Falling | Steeper |
These descriptions apply on intervals where the derivative stays on the stated side of zero. If \(f'\) crosses zero, analyze the portions on either side separately: the direction of \(f\) can change, and the behavior of the slope magnitude can change as well.
Worked Examples
Worked Example: Falling, Then Rising, While the Slope Changes
Let \(f(x)=x^2-4x\) on \([0,4]\). Differentiate to identify the derivative graph:
Evaluate the derivative at several inputs:
For \(0<x<2\), \(f'(x)<0\), so \(f\) is decreasing there. Its derivative values move from \(-4\) toward \(0\), so \(\lvert f'(x)\rvert\) decreases: \(f\) falls less steeply as \(x\) approaches \(2\). For \(2<x<4\), \(f'(x)>0\), so \(f\) is increasing. The derivative values move from \(0\) toward \(4\), so the rising function becomes steeper.
Check the derivative and the change in the function across the interval: \(f'(x)=2x-4\) as stated, and \(f(0)=0\), \(f(2)=4-8=-4\), \(f(4)=16-16=0\). The function falls from \(0\) to \(-4\), then rises from \(-4\) to \(0\), in agreement with the derivative signs.
Worked Example: Rising Then Falling as the Derivative Decreases
Let \(f(x)=4x-x^2\) on \([0,4]\). Its derivative is
The derivative graph is a decreasing straight line. Its values at selected inputs are:
On \(0<x<2\), the derivative is positive, so \(f\) rises. Its slope decreases from positive values toward zero, which means the rising graph becomes flatter. On \(2<x<4\), the derivative is negative, so \(f\) falls. Its values become more negative, and their magnitudes increase from zero, so the falling graph becomes steeper.
Check the derivative by differentiating \(4x-x^2\), which gives \(4-2x\). Also, \(f(0)=0\), \(f(2)=8-4=4\), and \(f(4)=16-16=0\). The function rises from \(0\) to \(4\), then falls to \(0\), matching the sign changes in \(f'\).
Worked Example: A Horizontal Tangent Without a Change from Rising to Falling
Let \(f(x)=\frac{(x-1)^3}{3}\) on \([0,3]\). Its derivative is
For example, \(f'(0)=1\), \(f'(1)=0\), \(f'(2)=1\), and \(f'(3)=4\). The derivative is nonnegative throughout the interval and positive except at \(x=1\). Therefore \(f\) is nondecreasing. Its slope magnitude falls from \(1\) to \(0\) as \(x\) approaches \(1\), then increases from \(0\) to \(4\) as \(x\) moves from \(1\) to \(3\). The tangent is horizontal at \(x=1\), but the derivative is positive on both sides, so the function rises on both sides.
Here we can check the stronger claim that \(f\) is strictly increasing. For any \(x<y\),
The last inequality holds because \(y-x>0\), and the quadratic factor is positive unless \(x=y=1\), which is impossible when \(x<y\). Finally, differentiating \(f(x)=\frac{(x-1)^3}{3}\) gives \((x-1)^2\), confirming the derivative used to read the trend.
A Common Misreading
Suppose a derivative graph rises from \(-4\) to \(-1\). It is tempting to say, “The derivative is rising, so the function is rising more steeply.” That conclusion confuses the direction in which \(f'\) changes with the size of the slope of \(f\).
Worked Example: An Increasing Derivative Can Mean a Flatter Falling Function
Take \(f(x)=\frac{x^2}{2}-5x\) on \([1,4]\). Then \(f'(x)=x-5\), so \(f'(1)=-4\) and \(f'(4)=-1\). The derivative increases across this interval, but remains negative. Thus \(f\) is falling. Meanwhile, \(\lvert f'(x)\rvert\) decreases from \(4\) to \(1\), so the falling function becomes flatter, not steeper.
Check by differentiating \(\frac{x^2}{2}-5x\), which gives \(x-5\). The function values are \(f(1)=\frac12-5=-\frac92\) and \(f(4)=8-20=-12\); the later value is smaller, confirming that the function falls.
Correction: First use the sign of \(f'\) to decide whether \(f\) rises or falls. Then compare the magnitudes of the derivative values to decide whether its slopes become steeper or flatter. An increasing derivative is not, by itself, evidence that \(f\) is rising.
A Reliable Reading Procedure
When interpreting a derivative graph over a specified input range, use the following sequence. Describe behavior interval by interval, especially if the graph meets or crosses the horizontal axis.
Identify where \(f'(x)=0\) or where the derivative graph crosses the horizontal axis. These points separate intervals that may have different signs.
Above the axis means \(f'>0\), so \(f\) rises; below the axis means \(f'<0\), so \(f\) falls.
Track the distance of the derivative graph from zero. Increasing distance means steeper tangent slopes; decreasing distance means flatter ones.
Say where each description applies, and note points where the derivative is zero. Do not infer a change from rising to falling just from a zero at one point.
What This Interpretation Does—and Does Not—Tell You
A derivative graph describes the local slopes of \(f\), so it gives reliable information about direction of change and slope magnitude when its values are known. It does not give the actual height of \(f\): adding a constant to \(f\) leaves its derivative unchanged. Nor does one derivative value determine a whole interval’s trend; interval conclusions rely on the derivative’s sign across that interval and the continuity and differentiability conditions stated above.
Be precise with words such as “steeper.” A positive slope becoming larger means a rising graph becomes steeper. A negative slope becoming larger means it becomes less negative, so a falling graph becomes flatter. Whenever the derivative is near or crosses zero, compare its distance from zero rather than relying only on whether its graph rises or falls.
Check Your Understanding
For each question, use the sign of the derivative to describe direction and its magnitude to describe steepness.
- If \(f'\) is negative and its values move from \(-6\) to \(-2\), what happens to \(f\), and how does its steepness change?
- If \(f'\) is positive and decreasing toward zero on an interval, does \(f\) rise or fall, and does it become steeper or flatter?
- What does \(f'(c)=0\) tell you about the tangent at \(c\)? Does it alone prove that \(f\) changes direction there?
- State the conditions needed to use the derivative sign test on \([a,b]\).
- If \(f'\geq 0\) throughout an interval, what can you conclude about \(f\)? What stronger conclusion follows if \(f'>0\) throughout its interior?
