Why Derivative Signs Can Prove Uniqueness
An equation may be difficult to solve explicitly, yet it can still be possible to show that it has no more than one solution on a specified interval. The key is to study how the function changes between any two proposed solutions.
The Mean Value Theorem makes that idea precise. If two different inputs give the same function value, then at some point between them the derivative must equal the secant slope between those inputs. That secant slope is zero. Therefore, a derivative that is always strictly positive or always strictly negative rules out two distinct inputs with the same output.
Precise Conditions
An interval may be open, closed, half-open, bounded, or unbounded. The argument applies to any two points \(a<b\) in the interval, provided the function is continuous on \([a,b]\) and differentiable on \((a,b)\). These are exactly the conditions needed to apply the Mean Value Theorem on that segment.
For the uniqueness result, the derivative must have the same strict sign at every point between any two points in the interval. For example, it is enough that \(f'(x)>0\) throughout the interior of the interval, or that \(f'(x)<0\) throughout that interior. The theorem then concerns solutions within the stated interval; it makes no claim about solutions outside it.
The Uniqueness Theorem
The same conclusion applies to \(f(x)=0\), which is often called finding the roots or zeros of \(f\). If the equation is \(f(x)=g(x)\), define \(h(x)=f(x)-g(x)\). The equation is then equivalent to \(h(x)=0\), so it suffices to show that \(h'\) has one strict sign on the interval.
Why the Argument Works
The proof is short, but it is important to see exactly where the hypotheses enter. Suppose, contrary to the claim, that \(f(x)=L\) has two distinct solutions \(u\) and \(v\) in \(I\). Relabel them so that \(u<v\). The segment \([u,v]\) lies in \(I\), and the assumptions allow the Mean Value Theorem to be applied there.
Let \(u<v\) and suppose \(f(u)=L\) and \(f(v)=L\).
There is a \(c\in(u,v)\) such that \(f'(c)=\frac{f(v)-f(u)}{v-u}\).
Since \(f(v)-f(u)=L-L=0\) and \(v-u>0\), the theorem gives \(f'(c)=0\).
This is impossible if \(f'\) is strictly positive everywhere or strictly negative everywhere in the interval.
Thus, there cannot be two distinct solutions. This proves the theorem. Notice that the proof does not calculate a solution; it rules out the possibility of two solutions by considering what the Mean Value Theorem would require if two existed.
Worked Example: A Polynomial Equation
Worked Example: A Polynomial Equation
Show that \(x^3+x=2\) has exactly one solution on \([0,2]\).
Define \(f(x)=x^3+x\). The equation becomes \(f(x)=2\). The function is a polynomial, so it is continuous on \([0,2]\) and differentiable on \((0,2)\). Its derivative is
For every \(x\in(0,2)\), \(x^2\geq 0\), so \(3x^2+1\geq 1>0\). The uniqueness theorem therefore shows that \(f(x)=2\) has at most one solution on \([0,2]\).
To establish that there is a solution, check \(x=1\):
So \(x=1\) is a solution, and the equation has at least one solution. Combined with the at-most-one result, this proves it has exactly one solution on the interval. Substitution verifies the claimed solution, while the derivative argument rules out every other one.
Worked Example: A Strictly Negative Derivative
Worked Example: A Strictly Negative Derivative
Show that \(\frac{1}{x+1}=\frac{1}{3}\) has exactly one solution on \([0,\infty)\).
Let \(f(x)=\frac{1}{x+1}\), which is defined and continuous on \([0,\infty)\), and differentiable in its interior. Differentiating gives
For every \(x>0\), the denominator \((x+1)^2\) is positive, so \(f'(x)<0\). For any two points \(a<b\) in \([0,\infty)\), the Mean Value Theorem gives a point \(c\in(a,b)\), where \(c>0\), and hence \(f'(c)<0\). The uniqueness theorem shows that \(f(x)=\frac{1}{3}\) has at most one solution on the interval.
Solve the equation directly:
The value \(2\) belongs to the interval. Check it in the original equation:
Thus \(x=2\) is a solution, and the at-most-one result proves it is the only solution on \([0,\infty)\).
Using the Difference of Two Functions
When an equation equates two expressions, neither expression needs to have a derivative of one strict sign by itself. Instead, subtract one side from the other. The sign of the derivative of that difference is what matters for uniqueness.
Worked Example: Comparing Two Polynomial Expressions
Show that \(x^2=2x+3\) has exactly one solution on \([2,\infty)\).
Set \(h(x)=x^2-(2x+3)=x^2-2x-3\). The original equation is equivalent to \(h(x)=0\). The function \(h\) is continuous on every closed segment in \([2,\infty)\) and differentiable in its interior. Its derivative is
For every \(x>2\), \(2x-2>4-2=2>0\). Consider any \(a<b\) in \([2,\infty)\). The Mean Value Theorem supplies \(c\in(a,b)\); in particular, \(c>2\), so \(h'(c)>0\). Therefore \(h(x)=0\), and hence the original equation, has at most one solution in the interval.
Test \(x=3\):
Both sides agree, so \(x=3\) is a solution. The uniqueness result proves it is the only solution on \([2,\infty)\). As a check using \(h\), \(h(3)=9-6-3=0\), as required.
At Most One Is Not the Same as Exactly One
A derivative-sign argument alone proves an upper bound on the number of solutions. It does not show that any solution exists. To conclude that an equation has exactly one solution, establish both parts:
- Use the derivative sign and the Mean Value Theorem to show that there is at most one solution.
- Find a solution directly, or use an appropriate existence theorem to show that at least one solution exists.
For instance, if \(f'\) is strictly positive on an interval, it does not follow that \(f(x)=100\) has a solution there. The function might never reach \(100\). The derivative condition prevents repeated values; it does not guarantee that a specified value is attained.
Common Pitfall: Checking the Derivative on Only Part of the Interval
The sign condition must hold throughout the region between any two possible solutions. It is not enough to check the derivative at one point, or to check it only on part of the interval.
Common Mistake: Ignoring a Change in Derivative Sign
A mistaken argument might claim that \(x^2=1\) has at most one solution on \([-2,2]\), because “the derivative of \(x^2\) is positive.” The derivative is \(2x\), but it is not positive throughout this interval: it is negative when \(x<0\), zero at \(x=0\), and positive when \(x>0\).
Indeed, substituting gives two distinct solutions:
The derivative-sign theorem cannot be applied on \([-2,2]\), because its strict-sign hypothesis fails. The corrected reasoning is not that the theorem gives uniqueness, but that there is no contradiction between the two solutions and the theorem: its required derivative condition was never satisfied.
A related caution is that a weak inequality such as \(f'(x)\geq 0\) is not the same hypothesis as \(f'(x)>0\). A weak sign by itself does not rule out a function being constant over part of an interval. For example, define \(f(x)=0\) for \(x\leq 0\) and \(f(x)=x^3\) for \(x>0\). This function is differentiable and has \(f'(x)\geq 0\), but every \(x\in[-1,0]\) satisfies \(f(x)=0\). The strict-sign condition in the theorem is a useful sufficient condition; it should not be silently weakened.
A Practical Proof Pattern
For an equation whose uniqueness you need to establish, organize the reasoning so that the domain, derivative, and conclusion are all explicit. This prevents common gaps, especially forgetting to restrict the claim to the interval on which the derivative sign is known.
State clearly which solutions are being counted.
For \(f(x)=L\), use \(f\). For \(f(x)=g(x)\), use \(h(x)=f(x)-g(x)\).
Verify continuity on each relevant closed segment and differentiability on its interior.
Show \(f'(x)>0\) everywhere relevant, or show \(f'(x)<0\) everywhere relevant.
Invoke the Mean Value Theorem uniqueness argument. If claiming exactly one, also verify existence.
Why This Method Is Useful
Uniqueness is valuable even when an explicit formula for a solution is unavailable. A derivative-sign proof can show that a numerical approximation or a solution found by another method refers to the only solution in the interval. It can also separate the work of proving existence from the work of proving uniqueness.
The essential reasoning is compact: two equal function values would force a zero derivative somewhere between them by the Mean Value Theorem. A derivative that is strictly positive or strictly negative cannot be zero there, so two such values are impossible. The interval and regularity assumptions matter because they are what make that theorem available.
Check Your Understanding
For each question, state the relevant interval and explain how the Mean Value Theorem supports the conclusion.
- If \(f'(x)<0\) throughout the interior of an interval, how many solutions can \(f(x)=L\) have there?
- Why does applying the Mean Value Theorem to two proposed solutions of \(f(x)=L\) produce a point where \(f'\) is zero?
- For an equation \(f(x)=g(x)\), which function should you differentiate to use this uniqueness argument?
- What additional fact is needed to upgrade “at most one solution” to “exactly one solution”?
- Why does the derivative of \(x^2\) not prove uniqueness of \(x^2=1\) on \([-2,2]\)?
