How Three Graphs Are Connected
A graph of \(f\) shows the function’s values. A graph of \(f'\) shows its tangent slopes, and a graph of \(f''\) shows how those slopes change. Matching the three graphs is therefore a consistency problem: each graph must agree with the derivative relationships encoded by the other two.
The vertical position of \(f\) alone does not determine either derivative. Adding a constant to \(f\) shifts its graph vertically but leaves \(f'\) and \(f''\) unchanged. By contrast, the sign and value of \(f'\) describe the slopes of \(f\), while the sign of \(f''\) describes the concavity of \(f\) and the direction in which \(f'\) changes.
Read the Graphs Interval by Interval
For a candidate graph of \(f\), inspect where its tangent slopes are positive, negative, or zero. A positive value \(f'(a)>0\) says that the tangent at \(x=a\) has positive slope. It does not, by itself, say that \(f\) is increasing throughout a neighborhood of \(a\). For that interval conclusion, establish that \(f'(x)>0\) throughout the interval in question. The corresponding statement holds for negative derivatives.
For a candidate graph of \(f'\), inspect its height and its own increasing or decreasing behavior. Where \(f'(x)>0\), the tangent slopes of \(f\) are positive; where \(f'(x)<0\), they are negative. Where \(f'\) is increasing, \(f''\geq 0\) wherever \(f''\) exists; where \(f'\) is decreasing, \(f''\leq 0\). Strictly positive (or negative) \(f''\) throughout an interval is sufficient for \(f'\) to be strictly increasing (or decreasing) there.
A candidate graph of \(f''\) gives a direct interval test for how \(f'\) changes. Positive \(f''\) on an interval means \(f'\) increases there, and negative \(f''\) means \(f'\) decreases there. These conclusions depend on the sign holding throughout the interval, not just at one point.
| Information on an interval | Required behavior |
|---|---|
| \(f'(x)>0\) | \(f\) is increasing on that interval |
| \(f'(x)<0\) | \(f\) is decreasing on that interval |
| \(f''(x)>0\) | \(f'\) is increasing; \(f\) is concave up |
| \(f''(x)<0\) | \(f'\) is decreasing; \(f\) is concave down |
A zero also carries useful information, but it must be interpreted carefully. If \(f'(a)=0\), the graph of \(f\) has a horizontal tangent at \(a\), assuming \(f'(a)\) exists. A local maximum or minimum requires further information, such as a change in the sign of \(f'\). If \(f''(a)=0\), that point is a candidate for a change in concavity, not a guarantee of one. Concavity changes only if the concavity on the two sides differs.
Why Interval Signs Give These Conclusions
The connection between derivative signs and graph behavior follows from the Mean Value Theorem. In the following argument, assume the relevant functions are continuous on each closed subinterval and differentiable on its interior, as required by that theorem.
For any \(a<b\) in an interval where the hypotheses hold, there is a \(c\in(a,b)\) such that \(f(b)-f(a)=f'(c)(b-a)\).
If \(f'(x)>0\) throughout the interval, then \(f'(c)>0\), and \(b-a>0\). Thus \(f(b)-f(a)>0\), so \(f(b)>f(a)\). This holds for every \(a<b\), proving that \(f\) is increasing.
If \(f''(x)>0\) throughout the interval, apply the Mean Value Theorem to \(f'\). For \(a<b\), \(f'(b)-f'(a)=f''(c)(b-a)>0\) for some \(c\in(a,b)\). Hence \(f'\) is increasing.
This argument explains why a sign on an interval is decisive: it controls every secant comparison within that interval. A derivative value at just one point supplies only the tangent slope at that point and does not provide the interval-wide condition used in the proof.
Worked Example: Match a Cubic with Its Two Derivatives
Worked Example: Match a Cubic with Its Two Derivatives
Suppose the candidate function is \(f(x)=x^3-3x\). Find the corresponding derivative graphs and describe the interval behavior that a matching set of sketches must show.
Differentiate once:
The zeros are \(x=-1\) and \(x=1\). The factors show that \(f'(x)>0\) when \(x<-1\), \(f'(x)<0\) when \(-1<x<1\), and \(f'(x)>0\) when \(x>1\). Therefore \(f\) increases on \((-\infty,-1)\), decreases on \((-1,1)\), and increases on \((1,\infty)\). Its graph has a local maximum at \(x=-1\) and a local minimum at \(x=1\), since the sign of \(f'\) changes from positive to negative and then from negative to positive.
The corresponding function values are
Differentiate again:
Thus \(f''(x)<0\) for \(x<0\), and \(f''(x)>0\) for \(x>0\). The graph of \(f\) is concave down to the left of zero and concave up to the right. The graph of \(f'\) decreases for \(x<0\) and increases for \(x>0\), so it has a minimum at \(x=0\). Also \(f(0)=0\), and the concavity changes there, giving an inflection point at \((0,0)\).
Check the proposed derivatives by differentiating the original function:
These calculations reproduce both candidate derivative formulas. A matching triple must therefore have a first-derivative graph crossing the horizontal axis at \(-1\) and \(1\), and a second-derivative graph crossing it at \(0\). Their signs must agree with the intervals just established.
Worked Example: Reconstruct a Function from Its Derivative
Worked Example: Reconstruct a Function from Its Derivative
Suppose a candidate graph for \(f'\) is described by \(f'(x)=x^2+x-2\), and the matching function is known to satisfy \(f(0)=3\). Determine \(f\), find \(f''\), and describe the graph behavior.
First factor the derivative:
Its zeros are \(-2\) and \(1\). The product is positive for \(x<-2\), negative for \(-2<x<1\), and positive for \(x>1\). So \(f\) increases, then decreases, then increases. The graph of \(f\) has a local maximum at \(-2\) and a local minimum at \(1\), because \(f'\) changes sign at each point.
To find the function, integrate the polynomial and include a constant:
Use the given value \(f(0)=3\):
Therefore,
Differentiate the candidate \(f'\) to obtain the second derivative:
This is negative for \(x<-\frac12\) and positive for \(x>-\frac12\). Thus \(f'\) decreases up to \(-\frac12\) and increases after \(-\frac12\), and \(f\) changes from concave down to concave up there. The value at that point is
For an additional graph check, the critical-point heights are
Finally, differentiate the reconstructed function:
Both derivatives agree with the proposed graphs. The additive constant was determined by the anchor \(f(0)=3\); without an anchor, the derivative graphs would not specify the vertical position of \(f\).
Worked Example: Use a Constant Second Derivative
Worked Example: Use a Constant Second Derivative
Suppose a candidate graph of \(f''\) is the horizontal line \(f''(x)=2\), and the other information is \(f'(0)=-2\) and \(f(0)=1\). Find all three formulas and check their graph behavior.
Since the derivative of \(2x\) is \(2\), integrating \(f''\) gives
Use \(f'(0)=-2\):
Thus \(f'(x)=2x-2\). Integrating again gives
The condition \(f(0)=1\) gives \(C_2=1\), so
The derivative \(f'(x)=2x-2\) is negative for \(x<1\), zero at \(1\), and positive for \(x>1\). Therefore \(f\) decreases up to \(x=1\), then increases, with a minimum value \(f(1)=0\). Since \(f''(x)=2>0\) everywhere, \(f'\) increases everywhere and \(f\) is concave up everywhere.
Check each formula by differentiation:
Also, \(f'(0)=-2\) and \(f(0)=1\), as required. The corresponding sketches must show a concave-up parabola, a rising straight-line graph for \(f'\), and a positive horizontal line for \(f''\).
A Common Matching Error
A tempting but incorrect shortcut is to treat a positive derivative at one point as proof of increase on an entire nearby interval. The statement \(f'(a)>0\) establishes the tangent slope at \(a\), not the signs of \(f'\) at other points.
For a concrete counterexample, define \(f(0)=0\) and, for \(t\ne0\), let
At zero, the difference quotient is
which tends to \(1\) as \(t\) tends to zero. Hence \(f'(0)=1>0\). But for \(t\ne0\),
Take \(t_n=1/\sqrt{2\pi n}\) for positive integers \(n\). Then \(\sin(1/t_n^2)=0\), \(\cos(1/t_n^2)=1\), and
for all sufficiently large \(n\). These points approach zero, so \(f'\) is not positive throughout any interval around zero. The correction is to use \(f'(a)>0\) only for the positive tangent slope at \(a\); to conclude that \(f\) increases on an interval, verify \(f'(x)>0\) throughout that interval.
A Practical Matching Procedure
When comparing several candidate sketches, use all three graphs together rather than relying on visual resemblance alone. Mark the important \(x\)-values—zeros, turning points, and points where a derivative is undefined—and compare the behavior on the intervals between them.
- Read the sign of \(f'\). Where it is positive or negative throughout an interval, check that \(f\) rises or falls there.
- Read the sign of \(f''\). Where it is positive or negative throughout an interval, check that \(f'\) rises or falls there, and that \(f\) is concave up or down.
- Check zeros with care. A zero of \(f'\) gives a horizontal tangent if the derivative exists; a local extremum requires a suitable sign change. A zero of \(f''\) does not by itself establish an inflection point.
- Check the derivative relationships. The slope of the \(f\)-graph at a point must agree with the value of the \(f'\)-graph there. The slope of the \(f'\)-graph must agree with the value of the \(f''\)-graph where that derivative exists.
- Allow for vertical shifts. If no function value is specified, different vertical placements of \(f\) can have the same derivative graphs.
A good match must satisfy every available check, not just one. For example, a proposed \(f\)-graph might have the right local maximum but the wrong concavity; a proposed \(f'\)-graph might have the right zeros but decrease where \(f''\) is positive. Any such contradiction rules out the triple.
Check Your Understanding
Use interval signs and derivative relationships to answer the following questions.
- If \(f'(x)<0\) throughout an interval, what must the graph of \(f\) do on that interval?
- If \(f''(x)>0\) throughout an interval, how does \(f'\) behave there, and what is the concavity of \(f\)?
- What does \(f'(a)>0\) tell you at the single point \(a\), and what additional information is needed to conclude that \(f\) increases on an interval?
- If \(f'(a)=0\), what can you conclude about the tangent at \(a\), and what further evidence is needed to classify a local extremum?
- Why does \(f''(a)=0\) alone fail to guarantee an inflection point?
