Why Two Function Values Need a Compatibility Check
Knowing a derivative determines a family of possible functions, not usually one specific function. On an interval, the family has the form \(F(x)=G(x)+C\), where \(G\) is one antiderivative of the given derivative and \(C\) is a constant.
A proposed function value such as \(F(a)=A\) selects one value of \(C\). If a second value \(F(b)=B\) is also proposed, it must select that same constant. The two conditions may each be possible on their own and still contradict one another when imposed together.
The Compatibility Criterion
Suppose \(g\) is the given derivative on an interval \(I\), and let \(G\) be a known antiderivative, so \(G'(x)=g(x)\) for every \(x\in I\). The antiderivative family on \(I\) is \(F(x)=G(x)+C\). For proposed conditions \(F(a)=A\) and \(F(b)=B\), substitute each point and value into this family:
There is one constant satisfying both conditions if and only if these two expressions are equal. Equivalently, the proposed values must satisfy:
The criterion can also be written as \(B-A=G(b)-G(a)\). This form says that the change between the two proposed function values must match the change in the known primitive between the two points.
Why Every Antiderivative Uses One Constant on an Interval
The compatibility test relies on the fact that any two antiderivatives of the same function on an interval differ by a constant. Here is a justification. Suppose \(F'(x)=g(x)\) and \(G'(x)=g(x)\) throughout \(I\), and define \(H(x)=F(x)-G(x)\). Then \(H'(x)=0\) on \(I\).
Take any two points \(x,y\in I\). Because \(I\) is an interval, every point between \(x\) and \(y\) also belongs to \(I\). The function \(H\) is continuous on the closed segment between them and differentiable in its interior. The Mean Value Theorem therefore gives a point \(c\) between \(x\) and \(y\) such that \(H(y)-H(x)=H'(c)(y-x)\). Since \(H'(c)=0\), the difference is zero, so \(H(y)=H(x)\). This holds for every pair \(x,y\in I\); hence \(H\) is constant on \(I\).
This also explains why the interval assumption matters: it lets us apply the argument between any two points in the domain. If the domain has separate components, points in different components need not share one constant. That issue is part of the earlier material on antiderivatives on disconnected domains; here, the compatibility test concerns two points in the same interval.
Worked Example: Two Compatible Values
Worked Example: Two Compatible Values
Suppose \(F'(x)=2x+3\) on \(\mathbb{R}\), and the proposed values are \(F(-1)=4\) and \(F(2)=16\). Determine whether one antiderivative satisfies both.
First find an antiderivative of the given derivative. Since \(\frac{d}{dx}(x^2+3x)=2x+3\), the family is \(F(x)=x^2+3x+C\).
Use the first condition. At \(x=-1\), the known part is \((-1)^2+3(-1)=1-3=-2\). Thus \(4=-2+C\), so \(C=6\).
Use the second condition independently. At \(x=2\), the known part is \(2^2+3(2)=4+6=10\). Thus \(16=10+C\), so \(C=6\).
Both conditions require the same constant, so they are compatible. The function is \(F(x)=x^2+3x+6\).
Check the result: \(F'(x)=2x+3\), \(F(-1)=1-3+6=4\), and \(F(2)=4+6+6=16\). The derivative and both proposed values are satisfied.
Worked Example: Each Value Works Alone, but Not Together
Worked Example: Each Value Works Alone, but Not Together
Suppose \(F'(x)=\cos x\) on \(\mathbb{R}\), with proposed conditions \(F(0)=2\) and \(F\left(\frac{\pi}{2}\right)=4\). Test compatibility.
An antiderivative of \(\cos x\) is \(\sin x\), because \(\frac{d}{dx}(\sin x)=\cos x\). Therefore \(F(x)=\sin x+C\).
The first condition gives \(2=\sin(0)+C=0+C\), so \(C=2\). The second condition gives \(4=\sin\left(\frac{\pi}{2}\right)+C=1+C\), so \(C=3\).
The conditions require different constants, \(2\) and \(3\). Therefore no single antiderivative satisfies both. Each value can be achieved separately by a suitable choice of \(C\), but they cannot both be achieved by the same choice.
For a direct check, the first condition forces \(F(x)=\sin x+2\), which gives \(F\left(\frac{\pi}{2}\right)=1+2=3\), not \(4\). The second condition instead forces \(F(x)=\sin x+3\), which gives \(F(0)=0+3=3\), not \(2\). Both candidate functions have derivative \(\cos x\), but neither meets both proposed values.
A Step-by-Step Test
A reliable calculation keeps the derivative, the primitive, and the proposed values distinct. The derivative specifies the family; each value then imposes a condition on its constant.
Choose \(G\) with \(G'=g\), where \(g\) is the given derivative.
Use \(F(x)=G(x)+C\) on the interval under consideration.
Substitute \(x=a\) and \(F(a)=A\) to obtain \(C=A-G(a)\).
Substitute \(x=b\) and \(F(b)=B\) to obtain \(C=B-G(b)\).
If the constants match, state the resulting function and check its derivative and both values. If they differ, state that the conditions are incompatible.
Worked Example: Values at Zero and One
Worked Example: Values at Zero and One
Suppose \(F'(x)=e^x+2x\) on \(\mathbb{R}\), and the proposed conditions are \(F(0)=0\) and \(F(1)=e\). Determine whether they are compatible.
Since \(\frac{d}{dx}(e^x+x^2)=e^x+2x\), the family is \(F(x)=e^x+x^2+C\).
At zero, the known part is \(e^0+0^2=1\). The condition \(F(0)=0\) gives \(0=1+C\), so \(C=-1\).
At one, the known part is \(e^1+1^2=e+1\). The condition \(F(1)=e\) gives \(e=e+1+C\), so \(C=-1\).
The constants agree, so the conditions are compatible and determine \(F(x)=e^x+x^2-1\).
Check the derivative: \(\frac{d}{dx}(e^x+x^2-1)=e^x+2x\). Check the values: \(F(0)=1+0-1=0\), and \(F(1)=e+1-1=e\). All three requirements hold.
A Common Mistake: Treating the Constants as Independent
A frequent error is to solve each condition correctly and then accept both answers without comparing them. That effectively allows one constant for the first value and a different constant for the second, even though the proposed function is a single member of the antiderivative family.
Worked Example: Correcting an Independent-Constants Error
Use \(F'(x)=\cos x\), \(F(0)=2\), and \(F\left(\frac{\pi}{2}\right)=4\). A mistaken calculation might write \(F(x)=\sin x+C_1\) for the first condition and \(F(x)=\sin x+C_2\) for the second, then conclude that both conditions can be met by choosing \(C_1=2\) and \(C_2=3\).
The error is that the problem asks for one function \(F\), so it must have one constant \(C\). Substituting the first condition gives \(C=2\); substituting the second gives \(C=3\). A single number cannot equal both \(2\) and \(3\).
Thus the conditions are incompatible. The correction is not to choose two constants, but to report that no one function in \(F(x)=\sin x+C\) satisfies both. Indeed, with \(C=2\), the second value is \(3\); with \(C=3\), the first value is \(3\). Each choice fails the other condition.
Choice of Primitive and the Role of the Domain
The compatibility conclusion does not depend on which particular primitive \(G\) you start with. If \(G_2(x)=G_1(x)+K\) for a constant \(K\), then the constant calculated from \(F(a)=A\) using \(G_2\) is \(A-G_2(a)=A-G_1(a)-K\). The constant calculated from \(F(b)=B\) also decreases by \(K\). Thus both required constants shift by the same amount, and whether they agree is unchanged.
Keep the domain in view. The criterion applies when \(a\) and \(b\) lie in the same interval on which the antiderivative family uses one constant. When the points lie in different components of a disconnected domain, the constants on those components can be chosen independently. In that case, a mismatch between the constants calculated on separate components is not, by itself, a contradiction.
For two proposed values at the same point, compatibility has an especially simple form: the proposed values must be equal. Substituting the same input into one function cannot produce two different outputs. More generally, for distinct points in one interval, compare the constants each value requires; do not infer compatibility merely because each condition can be satisfied separately.
Check Your Understanding
For each question, identify the antiderivative family first, then compare the constants required by the proposed values.
- If \(F'(x)=4x\), can \(F(0)=1\) and \(F(2)=5\) both hold? Show the constant from each condition.
- If \(F'(x)=\sin x\), can \(F(0)=3\) and \(F(\pi)=1\) both hold? Explain how you decide.
- State the compatibility condition for \(F(a)=A\) and \(F(b)=B\), using a known primitive \(G\) of \(F'\).
- Why is it necessary that the two points lie in the same interval for the single-constant criterion?
- If two different known primitives are used, why does the compatibility conclusion remain unchanged?
