From a Derivative Graph to a Function
A graph of \(f'\) does not show the values of \(f\). It shows how \(f\) changes: where \(f'\) is positive, \(f\) rises as \(x\) increases; where \(f'\) is negative, \(f\) falls; and where \(f'=0\) throughout an interval, \(f\) is constant there. Reconstructing a curve means using that information to build a compatible picture of \(f\).
There are two different levels of information a derivative graph can provide. Its sign determines intervals of increase and decrease, and sign changes can indicate possible extrema. Its size also matters: accumulated signed area gives the change in function value. But even a complete graph of \(f'\) does not specify the function’s vertical position unless at least one value of \(f\) is known.
Reading the Sign of the Derivative
On an interval where \(f\) is differentiable, a positive derivative means the function is increasing, and a negative derivative means it is decreasing. If the derivative is zero at every point of an interval, the function is constant there. These conclusions can be justified by the Mean Value Theorem: for any two points \(a<b\) in the interval, there is a point \(c\) between them with \(f(b)-f(a)=f'(c)(b-a)\).
Since \(b-a>0\), the sign of \(f'(c)\) determines the sign of \(f(b)-f(a)\). Thus, if \(f'>0\) throughout the interval, then \(f(b)>f(a)\); if \(f'<0\), then \(f(b)<f(a)\). If \(f'=0\) throughout, then \(f(b)=f(a)\). The theorem requires that \(f\) be continuous on the closed interval between the points and differentiable in its interior.
For a reconstruction, first mark where the derivative graph lies above, below, or on the \(x\)-axis. Use those intervals to sketch rising, falling, or level pieces of \(f\). At a point where the derivative changes sign, check whether the hypotheses for an extremum conclusion are satisfied.
Accumulating Changes from the Derivative Graph
The height of \(f\) changes according to the signed area under \(f'\). Area above the \(x\)-axis contributes a positive change; area below contributes a negative change. This is the Fundamental Theorem of Calculus in the form needed for reconstruction.
The identity explains how to use a derivative graph quantitatively. Start at a known function value, then add the signed area from that \(x\)-coordinate to the next one. The result is the new function value. Areas above and below the axis can cancel, so a curve may return to its earlier height even though it rose and fell in between.
Worked Example: A Maximum from a Linear Derivative Graph
Suppose \(f'(x)=2-2x\) for all real \(x\), and \(f(0)=3\). First find where the derivative is zero: \(2-2x=0\), so \(x=1\). For \(x<1\), \(2-2x>0\); for \(x>1\), \(2-2x<0\). Therefore \(f\) increases up to \(x=1\) and decreases after \(x=1\).
From \(x=0\) to \(x=1\), the derivative graph forms a triangle above the axis, with base \(1\) and height \(2\). Its signed area is \(\frac12(1)(2)=1\), so \(f(1)=f(0)+1=3+1=4\). From \(x=1\) to \(x=2\), the graph forms a triangle below the axis, with base \(1\) and height \(2\). Its signed area is \(-\frac12(1)(2)=-1\), so \(f(2)=f(1)-1=4-1=3\).
An antiderivative gives the full curve: integrating \(2-2x\) gives \(f(x)=2x-x^2+C\). Substituting \(f(0)=3\) gives \(3=C\), hence \(f(x)=2x-x^2+3\). Check: its derivative is \(2-2x\), as required; also \(f(0)=3\), \(f(1)=2-1+3=4\), and \(f(2)=4-4+3=3\). Since \(f(x)=4-(x-1)^2\), its value is at most \(4\) for every \(x\), so the maximum at \(x=1\) is global.
Extrema Require Care at the Sign Change
A positive-to-negative change in \(f'\) indicates that the function rises on one side and falls on the other. If \(f\) is also continuous at the point \(c\), this guarantees a local maximum there. A negative-to-positive change similarly guarantees a local minimum, again provided \(f\) is continuous at \(c\). Continuity is important when the derivative is undefined at \(c\): the behavior on either side does not, by itself, determine the value of \(f(c)\).
These sign-change tests are sufficient, not necessary. A function can have a non-strict local extremum without its derivative changing sign. For example, a constant function has both a local maximum and a local minimum at every interior point under the usual non-strict definitions, despite having derivative zero on both sides.
The sign-change conclusion follows from monotonicity. If \(f'>0\) immediately to the left of \(c\), then \(f\) is increasing there; continuity at \(c\) ensures the left-hand values do not exceed \(f(c)\). If \(f'<0\) immediately to the right, then \(f\) decreases there, so the right-hand values do not exceed \(f(c)\) either. Together these inequalities give a local maximum. Reversing the signs gives a local minimum.
Worked Example: A Minimum at a Point Where the Derivative Is Undefined
Suppose a derivative graph shows \(f'(x)=-2\) for \(x<0\) and \(f'(x)=1\) for \(x>0\), and we know \(f(-1)=4\). The derivative changes from negative to positive at \(0\), but is not defined there. To conclude that the point is a minimum, we also need continuity of \(f\) at \(0\). Consider the continuous function determined by these slopes and the anchor value.
From \(x=-1\) to \(x=0\), the derivative is \(-2\) over an interval of length \(1\), so the signed change is \((-2)(1)=-2\). Thus \(f(0)=4-2=2\). From \(x=0\) to \(x=2\), the derivative is \(1\) over a length of \(2\), so the change is \((1)(2)=2\). Thus \(f(2)=2+2=4\).
The compatible function is \(f(x)=2-2x\) for \(x\leq0\) and \(f(x)=2+x\) for \(x\geq0\). Check the anchor: \(f(-1)=2-2(-1)=4\). Check continuity: both pieces give \(f(0)=2\). Their derivatives are \(-2\) on the left and \(1\) on the right, matching the graph. For \(x<0\), \(2-2x>2\); for \(x>0\), \(2+x>2\). Therefore \(f(0)=2\) is a local minimum, even though \(f'\) is undefined at \(0\).
Why the Vertical Position Is Undetermined
If \(f\) is compatible with a derivative graph, then so is \(f+C\) for any constant \(C\), since \((f+C)'=f'\). Adding \(C\) moves the entire curve vertically without changing its slopes, intervals of increase or decrease, or the locations of its turning behavior. A derivative graph alone therefore cannot determine the function’s vertical position.
In fact, on an interval, any two differentiable functions with the same derivative differ by a constant. To see why, let \(f'=g'\) and define \(h=f-g\). Then \(h'=f'-g'=0\). For any two points \(a<b\) in the interval, the Mean Value Theorem applied to \(h\) gives \(h(b)-h(a)=h'(c)(b-a)=0\) for some \(c\) between them. Thus \(h(a)=h(b)\), so \(h\) is constant. One known value, such as \(f(a)\), fixes that constant and selects a particular vertical placement.
Worked Example: A Flat Minimum Interval and Its Heights
On the domain \([-2,2]\), suppose the derivative graph has value \(-2\) on \((-2,-1)\), value \(0\) on \((-1,1)\), and value \(2\) on \((1,2)\). Assume \(f\) is continuous at the joins and \(f(-2)=5\). From \(-2\) to \(-1\), the signed change is \((-2)(1)=-2\), so \(f(-1)=5-2=3\). From \(-1\) to \(1\), the derivative is zero, so the change is \(0(2)=0\), and \(f(1)=3\). From \(1\) to \(2\), the change is \((2)(1)=2\), so \(f(2)=3+2=5\).
One compatible formula is \(f(x)=1-2x\) on \([-2,-1]\), \(f(x)=3\) on \([-1,1]\), and \(f(x)=1+2x\) on \([1,2]\). Check the values at the joins: the left formula gives \(1-2(-1)=3\), the middle gives \(3\), and the right gives \(1+2(1)=3\). The endpoint values are \(1-2(-2)=5\) and \(1+2(2)=5\). Differentiating within each piece gives \(-2\), \(0\), and \(2\), respectively, matching the derivative graph.
The function decreases, stays level, and then increases. Every point of the flat segment \([-1,1]\) is a non-strict local minimum: nearby function values are never below \(3\), though some equal it. The exact heights here depend on the anchor \(f(-2)=5\). Without that anchor, adding any constant to the displayed function would preserve the derivative graph.
A Common Mistake: Ignoring Continuity
It is tempting to conclude that a positive-to-negative derivative sign change always means a local maximum. That claim can fail when the function is discontinuous at the sign-change point.
Worked Example: A Sign Change That Does Not Give a Maximum
Define \(f(x)=x\) for \(x<0\), \(f(0)=-1\), and \(f(x)=-x\) for \(x>0\). For \(x<0\), \(f'(x)=1>0\); for \(x>0\), \(f'(x)=-1<0\). A mistaken sign-chart conclusion would call \(x=0\) a local maximum.
But values close to \(0\) on either side are greater than \(-1\): if \(x<0\), then \(f(x)=x>-1\) whenever \(-1<x<0\); if \(x>0\), then \(f(x)=-x>-1\) whenever \(0<x<1\). Thus \(f(0)=-1\) is a local minimum, not a maximum. The function is not continuous at \(0\), because its limit there is \(0\), while \(f(0)=-1\). The correction is to check continuity at the sign-change point before applying the sign-change test.
A Practical Reconstruction Process
When sketching from a first-derivative graph, separate the qualitative shape from the numerical heights. If the graph gives exact derivative values, use signed areas and any known function value to calculate heights. If it gives only signs, sketch the rises, falls, and possible extrema, but do not invent precise function values or a unique vertical placement.
Identify where the graph is above, below, or on the \(x\)-axis, and note points where the derivative is zero or undefined.
Use positive, negative, and identically zero derivative intervals for increasing, decreasing, and constant behavior.
At a sign change, verify continuity at the point before concluding that a local extremum is guaranteed. Remember that a sign change is sufficient, not necessary.
Starting from a known function value, add each area above the axis and subtract each area below it to determine later heights.
If no function value is supplied, describe the family of vertically translated curves rather than claiming the graph has a unique height.
What the Reconstruction Can and Cannot Tell You
A first-derivative graph provides useful constraints, but those constraints should not be overstated. Sign information determines monotonic behavior on intervals. Sign changes can guarantee extrema when continuity is present. Signed areas determine changes in height when the derivative values and necessary hypotheses are available. A known point then pins down the vertical placement.
By contrast, without an anchor value, the curve may be shifted up or down by any constant. And if only the sign of \(f'\) is known, the exact accumulated heights may remain unknown. A careful sketch should show what follows from the derivative data while leaving undetermined features genuinely undetermined.
Check Your Understanding
Use derivative signs, signed changes, and continuity conditions to answer these questions.
- If \(f'(x)<0\) throughout an interval, what does this imply about \(f\) there, assuming the Mean Value Theorem’s hypotheses hold?
- A derivative graph has a positive-to-negative sign change at \(c\). What additional condition guarantees a local maximum at \(c\)?
- If the signed area under \(f'\) from \(a\) to \(b\) is \(-3\) and \(f(a)=7\), what is \(f(b)\)?
- Why can two different functions have exactly the same derivative graph?
- Can a non-strict local extremum occur even when \(f'\) does not change sign? Explain briefly.
