Reading a Derivative Graph
A graph labeled \(y=f'(x)\) shows the values of the derivative function, not the values of the original function \(f\). To find \(f'(a)\), locate the input \(a\) on the horizontal axis and read the vertical coordinate of the derivative graph directly above or below it. That coordinate is the derivative value.
This reading has a second interpretation. When \(f\) is differentiable at \(a\), \(f'(a)\) is the slope of the tangent to the graph of \(f\) at input \(a\). Thus a point on the graph of \(f'\) tells us the slope of \(f\) at the corresponding input.
Coordinates and Axes
A point on the derivative graph has the form \((x,f'(x))\). Its horizontal coordinate is an input to the derivative function; its vertical coordinate is the value of that derivative. For example, if the graph passes through \((2,3)\), then the point says \(f'(2)=3\). It does not say \(f(2)=3\).
Before reading a value, check what each axis represents and how it is scaled. Tick marks need not increase by one. If vertical tick labels increase by two, moving one vertical grid interval changes the derivative value by two. Similarly, the horizontal scale tells you which input is represented by a given vertical line.
After identifying the point above \(a\), its vertical coordinate gives the slope of \(f\) at \(a\). A positive coordinate means a positive slope, and a negative coordinate means a negative slope. The magnitude indicates the size of that slope: for instance, a value of \(-4\) represents a steeper negative slope than \(-1\). The graph of \(f'\) gives this information through its own vertical scale.
Why the Vertical Coordinate Is the Slope
The derivative function is defined by assigning to each input \(x\) the derivative value \(f'(x)\), wherever that derivative exists. Its graph is therefore the set of points \((x,f'(x))\). This explains why the vertical coordinate at input \(a\) is \(f'(a)\).
The connection between that value and the slope of \(f\) follows from the definition of a derivative. For a nonzero change \(h\), the quotient below is the slope of a secant line through the points on the graph of \(f\) at inputs \(a\) and \(a+h\). If \(f\) is differentiable at \(a\), the limit of those secant slopes is \(f'(a)\), the tangent slope at \(a\).
Consequently, reading \(f'(a)\) from the derivative graph is not a separate slope calculation from the shape of that graph. You read a vertical coordinate on the graph of \(f'\); that coordinate represents the slope of the different function \(f\) at input \(a\).
Worked Examples
Worked Example: Reading a Positive Derivative Value
Suppose a graph labeled \(y=f'(x)\) has horizontal tick marks at \(x=0,1,2,3\) and vertical tick marks at derivative values \(0,1,2,3,4\). The plotted graph passes through \((2,3)\). Find \(f'(2)\) and interpret it as a slope of \(f\).
Step 1: Locate the input. The requested input is \(a=2\), so locate \(x=2\) on the horizontal axis.
Step 2: Read the vertical coordinate. At \(x=2\), the graph is at height \(3\). Therefore the plotted point is \((2,3)\), and its vertical coordinate is \(3\).
Step 3: Interpret the value. Since \(f\) is differentiable at \(2\), the tangent slope of \(f\) at input \(2\) is \(3\). The positive sign indicates a positive slope.
Check: The graph’s point has horizontal coordinate \(2\) and vertical coordinate \(3\). In \((x,f'(x))\), that means \(f'(2)=3\), as read above.
Worked Example: Reading a Negative Value Between Grid Points
A graph of \(y=f'(x)\) is a straight line segment from \((1,-2)\) to \((3,-6)\). Find \(f'(2)\), then state the slope of \(f\) at \(2\). The horizontal spacing from \(x=1\) to \(x=3\) covers two equal intervals.
Step 1: Locate the input within the segment. The input \(2\) is halfway between \(1\) and \(3\), since \(2-1=1\) and \(3-2=1\).
Step 2: Find the graph’s height there. Along a straight segment, halfway between the endpoints’ vertical coordinates is halfway between \(-2\) and \(-6\). The vertical change from \(-2\) to \(-6\) is \(-6-(-2)=-4\). Half of that change is \(-4/2=-2\). Starting at \(-2\), the height at \(x=2\) is \(-2+(-2)=-4\).
Step 3: Interpret the value. The tangent slope of \(f\) at \(2\) is \(-4\). It is negative, and its magnitude is \(4\).
Check: The segment changes from \(-2\) to \(-6\), a total vertical change of \(-4\) over two equal horizontal intervals. Each interval changes the graph’s height by \(-4/2=-2\). Thus at the midpoint \(x=2\), the height is \(-2-2=-4\), confirming the reading.
Worked Example: Using a Non-Unit Vertical Scale
A graph of \(y=f'(x)\) has horizontal tick labels \(0,1,2\). On the vertical axis, successive grid lines are labeled \(0\), \(0.5\), \(1.0\), and \(1.5\). The plotted point above \(x=-1\) is on the grid line labeled \(1.5\). Find \(f'(-1)\) and interpret it.
Step 1: Identify the input. The point is above the horizontal-axis location \(x=-1\).
Step 2: Use the vertical labels. The relevant grid line is labeled \(1.5\). The derivative value is therefore \(1.5\), or \(3/2\). It is not determined by counting grid intervals as though each interval represented one full unit: the labeled scale shows that each vertical interval is \(0.5\).
Step 3: Interpret the value. The tangent slope of \(f\) at input \(-1\) is \(3/2\), which is positive.
Check: The point’s coordinates are \((-1,1.5)\). Since a point on the derivative graph is of the form \((x,f'(x))\), its coordinates give \(f'(-1)=1.5\), confirming the result.
A Reliable Reading Process
For a value at a specified input, use the graph’s axes and coordinates in a consistent order. This is especially useful when the graph has a non-unit scale or the requested input does not lie at a labeled tick.
Check that the plotted graph represents \(y=f'(x)\), not \(y=f(x)\) or another quantity.
Find \(a\) on the horizontal axis, using the horizontal tick labels to determine its position.
Follow the graph at \(x=a\) to the vertical axis scale. That vertical coordinate is \(f'(a)\).
Use the sign and magnitude of \(f'(a)\) to describe the slope of \(f\) at \(a\), then check that you used the vertical—not horizontal—coordinate.
Points, Segments, and Missing Values
A plotted point gives a derivative value at its horizontal coordinate. If the graph includes a straight segment, a value at an input between its endpoints can be read from the segment’s height. When the input is halfway between the endpoints horizontally, the height is halfway between their vertical coordinates. For other positions on a straight segment, use the same proportional change along the segment, while respecting both axis scales.
Do not assume that a graph supplies a value where it does not show one. For example, an open circle at \(x=a\), with no filled point at that same input, indicates that the displayed graph does not assign the value shown by the open circle to \(a\). If the graph has a gap or does not cover \(a\), it does not provide a derivative value there. A graph’s absence of a plotted value is not itself permission to fill one in.
Also distinguish the location of a point from its height. For a derivative graph, the \(x\)-coordinate identifies where to evaluate \(f'\), and the \(y\)-coordinate gives the resulting derivative value. Switching those roles changes the question.
Common Mistake: Reading the Wrong Coordinate
Suppose a graph of \(y=f'(x)\) contains the point \((4,-2)\), and the question asks for \(f'(4)\). A common mistake is to use the first coordinate and claim \(f'(4)=4\). But \(4\) is the input, not the derivative value.
The correction is to read the coordinates in order: \((x,f'(x))=(4,-2)\). Therefore \(f'(4)=-2\), so the tangent slope of \(f\) at \(4\) is negative two. The wrong answer used the horizontal coordinate as if it were the vertical coordinate.
What the Reading Tells You
Every point on a derivative graph links an input to a slope of the original function. A point above the horizontal axis has a positive derivative value, so the corresponding tangent slope of \(f\) is positive. A point below the axis has a negative derivative value, so that slope is negative. A larger absolute vertical coordinate represents a slope with larger magnitude.
Keep the conclusion local to the input you read: the point at \(x=a\) gives the slope of \(f\) at \(a\). The vertical coordinate does not give the height of the graph of \(f\), and one point on \(f'\) does not specify the full graph of \(f\). It gives a precise slope value at the matching input.
Check Your Understanding
Use the coordinate meaning of a derivative graph to answer each question.
- A graph of \(y=f'(x)\) contains \((3,5)\). What is \(f'(3)\), and what does this value represent for \(f\)?
- A derivative graph contains the point \((-2,-1.5)\). State the derivative value at \(-2\) and its sign.
- A straight segment of the graph of \(f'\) joins \((0,2)\) to \((4,6)\). What is \(f'(2)\)?
- On a graph labeled \(y=f'(x)\), what information does the horizontal coordinate of a point give? What does the vertical coordinate give?
- At \(x=a\), the graph shows an open circle and no filled point. Can you read the derivative value from that open circle as the value at \(a\)? Explain.
