From Derivative Graphs to Function Shape
A graph of \(f'\) is not a graph of \(f\): its height at a point gives the slope of \(f\) there. A graph of \(f''\) gives information about how those slopes change. Reading both graphs together lets us reconstruct the function’s main shape without knowing its formula.
The key is to keep three questions separate. Is \(f\) rising or falling? Is its graph bending upward or downward? And where is the graph vertically positioned? The first two are answered by derivatives; the last generally requires a known value of \(f\).
What Each Derivative Tells You
If the graph of \(f'\) lies above the \(x\)-axis, then \(f'(x)>0\), so \(f\) is increasing there. If it lies below the axis, then \(f'(x)<0\), so \(f\) is decreasing. A zero of \(f'\) is a possible local maximum or minimum: check the signs on both sides to determine whether \(f\) changes direction.
The graph of \(f''\) gives concavity directly. Alternatively, if you are given only the graph of \(f'\), use its slope: where \(f'\) is increasing, \(f''>0\) and \(f\) is concave up; where \(f'\) is decreasing, \(f''<0\) and \(f\) is concave down. A horizontal portion of \(f'\) means \(f''=0\) there, so the second-derivative test alone does not label that portion as strictly concave up or down.
A candidate inflection point occurs where concavity might change. The test is whether concavity actually changes across the point—not merely whether \(f''=0\), or whether \(f''\) is undefined at that point. The earlier material on inflection points and concavity gives the detailed criteria.
Why the Sign Rules Work
The sign of \(f'\) determines whether function values rise or fall. Suppose \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\). By the Mean Value Theorem, some \(c\) between \(a\) and \(b\) satisfies the following equation. If \(f'\) is positive throughout the interval, the right-hand side is positive, so \(f(b)>f(a)\). The same argument with a negative derivative gives a decrease.
The concavity rule follows by applying the Mean Value Theorem to \(f'\). If \(f''>0\) throughout an interval, then \(f'\) is increasing there. Thus the slopes of \(f\) increase, which is the defining behavior of concavity up. If \(f''<0\), the slopes decrease, giving concavity down. These sign conclusions require the stated derivatives to exist on the intervals being considered; check points where a derivative is zero or undefined separately.
Worked Example: Combine Polynomial Derivative Evidence
Worked Example: Combine Polynomial Derivative Evidence
Suppose the derivative graphs have the equations \(f'(x)=(x+1)(x-2)\) and \(f''(x)=2x-1\), and the function passes through \((0,1)\). Determine the intervals of increase, decrease, and concavity, and list the main points needed for a sketch.
First find where \(f'\) is zero:
Testing the intervals gives \(f'(x)>0\) for \(x<-1\), \(f'(x)<0\) for \(-1<x<2\), and \(f'(x)>0\) for \(x>2\). Therefore \(f\) increases on \((-\infty,-1)\), decreases on \((-1,2)\), and increases on \((2,\infty)\). The sign changes identify a local maximum at \(x=-1\) and a local minimum at \(x=2\).
Next, solve \(f''(x)=0\):
Since \(2x-1<0\) for \(x<\frac12\) and \(2x-1>0\) for \(x>\frac12\), \(f\) is concave down on \((-\infty,\frac12)\) and concave up on \((\frac12,\infty)\). The concavity changes at \(x=\frac12\).
To find the function values, integrate \(f'\) and use the given point:
From \(f(0)=1\), we get \(C=1\). Substitution gives \(f(-1)=\frac{13}{6}\), \(f(2)=-\frac73\), and \(f(\frac12)=-\frac1{12}\). So the sketch has a local maximum at \((-1,\frac{13}{6})\), a local minimum at \((2,-\frac73)\), and a change in concavity at \((\frac12,-\frac1{12})\).
Check the result by differentiating the reconstructed function: \(f'(x)=x^2-x-2=(x+1)(x-2)\), and \(f''(x)=2x-1\). Both derivatives match the supplied evidence, and \(f(0)=1\), as required.
Reconstructing Heights from a Derivative Graph
Sign information tells you the direction of change, but not how far the function moves. If \(f'\) is given as a graph, the change in \(f\) between two points is the signed area under that graph. Areas above the axis contribute positively; areas below contribute negatively. This is the Fundamental Theorem of Calculus, provided \(f'\) is integrable on the interval.
A known value such as \(f(a)\) therefore anchors the vertical position. Without one, derivative evidence determines the shape only up to an additive constant: moving a candidate graph straight up or down leaves both its derivatives unchanged.
Worked Example: Use an Anchor to Reconstruct a Function
Worked Example: Use an Anchor to Reconstruct a Function
Suppose the graph of \(f'\) is the curve \(f'(x)=(x+1)e^{-x}\), the graph of \(f''\) is \(f''(x)=-xe^{-x}\), and \(f(0)=2\). Determine the main shape features and reconstruct \(f\).
Because \(e^{-x}>0\) for every real \(x\), the sign of \(f'\) is the sign of \(x+1\). Thus \(f\) decreases on \((-\infty,-1)\), increases on \((-1,\infty)\), and has a local minimum at \(x=-1\). The second derivative is positive for \(x<0\) and negative for \(x>0\). Hence \(f\) is concave up on \((-\infty,0)\), concave down on \((0,\infty)\), and changes concavity at \(x=0\).
Integrate \(f'\). Since the derivative of \(-(x+2)e^{-x}\) is \((x+1)e^{-x}\),
Use the anchor \(f(0)=2\): \(-2+C=2\), so \(C=4\). The reconstructed function is \(f(x)=4-(x+2)e^{-x}\).
The minimum value is \(f(-1)=4-e\), and the concavity change occurs at \((0,2)\). Check by differentiating: \(f'(x)=(x+1)e^{-x}\), and differentiating again gives \(f''(x)=-xe^{-x}\). Also, \(f(0)=4-2=2\). The function and both derivative graphs agree with all the given evidence.
A First-Derivative Graph Can Show Concavity Changes
Sometimes the graph of \(f'\) is easier to read than a separate graph of \(f''\). The slope of \(f'\) shows the sign of \(f''\), while the location of \(f'\) relative to the horizontal axis shows the sign of \(f'\). These are different pieces of information: a positive value of \(f'\) does not mean that \(f'\) is increasing.
Worked Example: Read Both Signs and Slopes from a Graph
Worked Example: Read Both Signs and Slopes from a Graph
Suppose, on the real line, the graph of \(f'\) consists of the two line segments \(f'(x)=x+1\) for \(x\leq1\) and \(f'(x)=3-x\) for \(x\geq1\). Also suppose \(f(0)=0\). Describe the shape of \(f\) and reconstruct it.
The first segment crosses the \(x\)-axis at \(x=-1\); the second crosses at \(x=3\). Thus \(f'<0\) for \(x<-1\), \(f'>0\) for \(-1<x<3\), and \(f'<0\) for \(x>3\). So \(f\) decreases, then increases, then decreases. It has a local minimum at \(x=-1\) and a local maximum at \(x=3\).
The slope of the first segment of \(f'\) is \(1\), so \(f''=1\) for \(x<1\) and \(f\) is concave up there. The slope of the second segment is \(-1\), so \(f''=-1\) for \(x>1\) and \(f\) is concave down there. The graph of \(f'\) has a corner at \(x=1\), so \(f''(1)\) is undefined; nevertheless, concavity changes from up to down, making \(x=1\) an inflection point.
Integrate on each side of \(1\). For \(x\leq1\), \(f'(x)=x+1\), so \(f(x)=\frac{x^2}{2}+x+C_1\). The condition \(f(0)=0\) gives \(C_1=0\), hence \(f(1)=\frac32\). For \(x\geq1\), integration gives \(f(x)=3x-\frac{x^2}{2}+C_2\). Matching the function value at \(1\) requires \(\frac52+C_2=\frac32\), so \(C_2=-1\).
Therefore \(f(x)=\frac{x^2}{2}+x\) for \(x\leq1\), and \(f(x)=3x-\frac{x^2}{2}-1\) for \(x\geq1\). The extremum values are \(f(-1)=-\frac12\) and \(f(3)=\frac72\), and \(f(1)=\frac32\). Check: differentiating the two formulas gives \(x+1\) and \(3-x\), respectively; both give \(f(1)=\frac32\). The first derivatives also agree at \(1\), with value \(2\), as the supplied graph requires.
A Common Mistake: Confusing Height with Concavity
A positive value of \(f'\) means that \(f\) is increasing; it does not say whether \(f\) is concave up or down. For the previous example, at \(x=2\), \(f'(2)=1>0\), while the slope of the graph of \(f'\) is \(-1\). Therefore \(f\) is increasing but concave down near \(x=2\).
A mistaken conclusion would be: “The \(f'\) graph is above the axis, so \(f''>0\), and \(f\) is concave up.” The correction is to read height and slope separately: height above the axis gives \(f'>0\); a decreasing \(f'\) gives \(f''<0\). In a sketch, record these two sign patterns in separate rows before combining them.
Check Your Understanding
Use sign changes and derivative-graph slopes carefully; distinguish what the evidence determines from what requires an anchor value.
- If \(f'\) is negative to the left of \(x=2\) and positive to the right, what type of local extremum occurs at \(x=2\)?
- If \(f'\) is decreasing throughout an interval, what does that imply about the concavity of \(f\) there?
- If \(f'(x)>0\) and \(f''(x)<0\) throughout an interval, describe both the direction of change and the concavity of \(f\).
- Why can two functions with identical first derivatives differ in their vertical positions?
- If \(f'(x)=0\) at a point, what additional evidence is needed to decide whether \(f\) has a local extremum there?
