From Zero Slope to a Constant Function
A derivative measures the instantaneous rate of change of a function. If that rate is zero at one point, the function has a horizontal tangent there, but this alone says little about its values elsewhere. A stronger condition is that the derivative be zero at every interior point of an interval. The Mean Value Theorem then shows that no two function values on that interval can differ.
This result turns information about a derivative into a conclusion about the entire function. It also provides a useful way to determine unknown constants: if two functions have the same derivative throughout an interval, their difference is constant there.
For an interval consisting of a single point, the conclusion is immediate. The proof below treats intervals containing at least two points. Continuity at any included endpoint is part of the hypothesis; differentiability at the endpoints is not required.
Why the Mean Value Theorem Gives Constancy
The Mean Value Theorem states that if a function is continuous on a closed interval \([a,b]\) and differentiable on its interior \((a,b)\), then there exists a point \(c\in(a,b)\) such that
To use this theorem, take any two points \(x,y\in I\) with \(x<y\). Because \(I\) is an interval, every point between \(x\) and \(y\) also belongs to \(I\). Thus \(f\) is continuous on \([x,y]\) and differentiable on \((x,y)\), as required.
Let \(x,y\in I\) with \(x<y\). The goal is to show their function values are equal.
There is some \(c\in(x,y)\) for which \(f(y)-f(x)=f'(c)(y-x)\).
Since \(c\in I\) is an interior point and \(f'(c)=0\), it follows that \(f(y)-f(x)=0\).
Thus \(f(y)=f(x)\). Since the two points were arbitrary, all values of \(f\) on \(I\) are equal.
In equations, the key step is
If the points are presented in the opposite order, exchange their names and apply the same argument. If they are equal, their function values are already equal. Therefore every pair of points in \(I\) has the same function value, which is exactly what it means for \(f\) to be constant on \(I\).
Using One Known Value
The theorem identifies the shape of the function: it is constant. A single known value then identifies which constant it is. This is useful even when the function is specified only by conditions on its derivative rather than by an explicit formula.
Worked Example: Determine the Function from Its Derivative and One Value
Suppose \(f\) is continuous on \([-2,4]\), differentiable on \((-2,4)\), and \(f'(x)=0\) for every \(x\in(-2,4)\). If \(f(3)=7\), find \(f(-2)\) and \(f(4)\).
For any two points in \([-2,4]\), the theorem just proved gives equal function values. In particular, take the points \(3\) and \(-2\): \(f(-2)=f(3)=7\). Taking the points \(3\) and \(4\) gives \(f(4)=f(3)=7\).
Therefore, \(f(x)=7\) for every \(x\in[-2,4]\). Check: this constant function has derivative \(0\) on \((-2,4)\) and takes the required value \(f(3)=7\).
Equal Derivatives Mean a Constant Difference
The zero-derivative theorem also compares two functions. Suppose \(f\) and \(g\) are both continuous on an interval \(I\) and differentiable at every interior point. Define their difference by \(h(x)=f(x)-g(x)\). If \(f'(x)=g'(x)\) at every interior point, then
Applying the theorem to \(h\) shows that \(h\) is constant. Thus \(f(x)-g(x)=C\) throughout \(I\), or equivalently \(f(x)=g(x)+C\). A known value at one point can determine \(C\).
Worked Example: Find Unknown Polynomial Coefficients
Let \(P(x)=x^3+ax^2+bx+c\) and \(Q(x)=x^3+2x^2-5x+7\). Suppose \(P'(x)=Q'(x)\) for every \(x\in(-1,3)\), and \(P(1)=4\). Find \(a\), \(b\), and \(c\), then determine \(P(x)-Q(x)\).
Differentiate each polynomial:
Equality of the derivatives implies \(2ax+b=4x-5\) for every \(x\in(-1,3)\). Rearrange this as \((2a-4)x+(b+5)=0\). Since \(0\) and \(1\) both belong to \((-1,3)\), substitute each value. At \(x=0\), \(b+5=0\), so \(b=-5\). At \(x=1\), \((2a-4)+(b+5)=0\). Substituting \(b=-5\) gives \(2a-4=0\), so \(a=2\).
Now use \(P(1)=4\): \(1+a+b+c=4\). Substituting \(a=2\) and \(b=-5\) gives \(1+2-5+c=4\), so \(c=6\). Therefore \(P(x)=x^3+2x^2-5x+6\), and
Check: \(P'(x)=3x^2+4x-5=Q'(x)\), and \(P(1)=1+2-5+6=4\). The difference is indeed the constant \(-1\).
Worked Example: Determine a Constant Difference from a Trigonometric Condition
Suppose \(F(x)=\sin x+ax+b\) and \(G(x)=\sin x+2x-1\) have equal derivatives for every real \(x\), and \(F(0)=4\). Find \(a\), \(b\), and \(F(x)-G(x)\).
Differentiate: \(F'(x)=\cos x+a\) and \(G'(x)=\cos x+2\). Equality for every \(x\) gives \(\cos x+a=\cos x+2\), hence \(a=2\). The value at zero is \(F(0)=\sin 0+2\cdot0+b=b\), so \(b=4\).
Consequently, \(F(x)=\sin x+2x+4\), and subtracting \(G\) gives \(F(x)-G(x)=5\) for every real \(x\). Check: \(F'(x)=\cos x+2=G'(x)\), and \(F(0)=4\). Also, direct subtraction gives \((\sin x+2x+4)-(\sin x+2x-1)=5\), as required.
Conditions Matter
Both parts of the hypothesis are essential to the reasoning. The derivative must vanish at every interior point, not merely at a selected point. Also, the domain must be an interval so that the Mean Value Theorem can be applied between any two points in it.
Worked Example: A Zero Derivative at One Point Is Not Enough
Consider \(f(x)=x^2\) on \([-1,1]\). Its derivative is \(f'(x)=2x\), so \(f'(0)=0\). However, \(f(0)=0\) and \(f(1)=1\), which are not equal. Thus \(f\) is not constant.
There is no contradiction with the theorem: the derivative is zero at \(x=0\) only, not throughout \((-1,1)\). In particular, the Mean Value Theorem between \(0\) and \(1\) gives some \(c\in(0,1)\) with \(\frac{f(1)-f(0)}{1-0}=f'(c)\). Here the left side is \(1\), and \(f'(c)=2c=1\) when \(c=\frac12\). The theorem's conclusion is consistent with the nonzero change in function value.
An interval condition also prevents a separate problem. Define \(u(x)=0\) for \(x<0\) and \(u(x)=1\) for \(x>0\), with domain \((-\infty,0)\cup(0,\infty)\). At every point of this domain, \(u\) is locally constant, so \(u'(x)=0\). Yet it is not constant on its whole domain: values on the negative component are \(0\), while values on the positive component are \(1\). The domain is not an interval, and the Mean Value Theorem cannot connect a negative point to a positive one within that domain.
Why This Result Is Useful
Knowing a derivative can leave one piece of information undetermined: an additive constant. The result explains precisely why. If two candidate functions have the same derivative on an interval, their difference has zero derivative there, so the candidates can differ only by a constant. A value at one point fixes that constant.
The same idea supports proofs of identities. When two expressions have equal derivatives on an interval, comparing their values at one convenient point can establish their relationship everywhere on that interval. The logic always depends on checking the hypotheses: the functions must be continuous on the relevant interval, differentiable in its interior, and have equal derivatives throughout that interior.
Check Your Understanding
For each question, identify which hypothesis or conclusion of the zero-derivative theorem is relevant.
- If \(f'(x)=0\) for every interior point of an interval and \(f(2)=-3\), what must \(f(x)\) equal throughout that interval?
- Suppose \(f\) and \(g\) are continuous on \([0,5]\), differentiable on \((0,5)\), and \(f'(x)=g'(x)\) throughout \((0,5)\). What can you conclude about \(f-g\)?
- For \(p(x)=x^2+ax+b\), suppose \(p'(x)=2x+6\) throughout an interval containing \(0\) and \(1\). Determine \(a\).
- Why does \(f'(0)=0\) alone fail to show that \(f\) is constant? Give a specific example.
- Why can the zero-derivative theorem not be applied across both components of \((-\infty,0)\cup(0,\infty)\) as though that domain were an interval?
