Solve a Multi-Theorem Real Analysis Problem

Turn One Problem Into Several Smaller Goals

A multi-theorem proof is not a contest to mention as many results as possible. The goal is to identify which fact answers each part of the question, then connect those facts without assuming more than they provide. In the previous tutorial, the central task was to test whether a conclusion survives when a hypothesis is removed. Here we take the complementary approach: given sufficient hypotheses, we assemble established theorems into a useful conclusion.

Consider a continuous function on a closed interval whose values have opposite signs at the endpoints and whose derivative is bounded below by a positive constant. The endpoint signs suggest the Intermediate Value Theorem: they can give existence of a zero. A positive derivative suggests the Strict Monotonicity from a Positive Derivative theorem: it can give uniqueness. But these two conclusions do not yet say how close an approximate zero is to the actual one. For that, the Mean Value Theorem turns the derivative bound into a distance estimate.

The proof has three distinct jobs. Keeping them separate helps prevent a common mistake: using an existence theorem to claim uniqueness, or using strict monotonicity to claim a quantitative error bound that it does not supply by itself.

Find a zero.
Use continuity and the endpoint signs with the Intermediate Value Theorem.

Show the zero is unique.
Use a strictly positive derivative and the Strict Monotonicity from a Positive Derivative theorem.

Estimate distance from the zero.
Apply the Mean Value Theorem between the zero and a test point, using the derivative lower bound.

The Combined Root and Error Theorem

The following result packages those jobs into a single statement. The positive constant $m$ is a lower bound for the slope: wherever the derivative exists, the function increases at a rate of at least $m$. The resulting estimate says that a small value of the function forces the input to be close to its zero.

Theorem (Unique Root with a Residual Error Bound): Let $a<b$, and let $f:[a,b]\to\mathbb R$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $f(a)<0<f(b)$, and suppose there is an $m>0$ such that $f'(x)\geq m$ for every $x\in(a,b)$. Then $f$ has exactly one zero $c\in(a,b)$. Moreover, for every $x\in[a,b]$, $$ |x-c|\leq \frac{|f(x)|}{m}. $$

Proof: Since $f$ is continuous on $[a,b]$ and $0$ lies strictly between $f(a)$ and $f(b)$, the Intermediate Value Theorem gives a point $c\in(a,b)$ with $f(c)=0$. The derivative satisfies $f'(x)\geq m>0$ at every interior point. By the Strict Monotonicity from a Positive Derivative theorem, $f$ is strictly increasing on $[a,b]$. A strictly increasing function cannot take the value zero at two distinct points, so $c$ is the unique zero.

It remains to prove the estimate. If $x=c$, both sides are zero. Suppose $x\neq c$. The function is continuous on the closed interval whose endpoints are $x$ and $c$, and differentiable in its interior, so the Mean Value Theorem applies there. It gives a point $\xi$ strictly between $x$ and $c$ such that $$ f(x)-f(c)=f'(\xi)(x-c). $$ Since $f(c)=0$ and $f'(\xi)\geq m$, we have $$ |f(x)|=f'(\xi)|x-c|\geq m|x-c|. $$ Dividing by $m>0$ proves the claimed estimate. The argument also covers $x=a$ and $x=b$: in either case, the point $\xi$ supplied by the Mean Value Theorem lies in $(a,b)$, where the derivative bound holds. $\square$

Notice how the hypotheses are used. Continuity on the entire closed interval and the endpoint signs provide existence. Differentiability on the open interval and the positive derivative provide uniqueness and the estimate. The inequality $m>0$ matters twice: it implies strict increase and makes division by $m$ valid.

Worked Example: Certifying an Approximation to a Polynomial Root

Let $f(x)=x^2+x-1$ on $[0,1]$. This function is continuous on the interval and differentiable on its interior. Its endpoint values are $f(0)=-1<0$ and $f(1)=1>0$, while $$ f'(x)=2x+1\geq 1 \qquad (0<x<1). $$ The theorem applies with $m=1$, so $f$ has exactly one zero $c\in(0,1)$.

Take $x=3/5$ as a proposed approximation. Direct substitution gives $$ f(3/5)=\frac{9}{25}+\frac{15}{25}-\frac{25}{25}=-\frac{1}{25}. $$ Therefore the error estimate gives $$ \left|\frac35-c\right|\leq \frac{|f(3/5)|}{1}=\frac{1}{25}. $$ No explicit formula for $c$ is needed to certify this accuracy. The function value at the test point and the slope bound suffice.

How Function Perturbations Move a Root

The residual estimate has another use. If a point is a zero of a nearby function, then it is an approximate zero of the original function. The uniform distance between the functions therefore gives a bound on the distance between their roots. This is a stability result: it quantifies how sensitive the root is to changes in the function.

Theorem (Stability of a Root Under Uniform Perturbation): Let $f:[a,b]\to\mathbb R$ be continuous on $[a,b]$, differentiable on $(a,b)$, and suppose $f'(x)\geq m>0$ for every $x\in(a,b)$. Let $c\in[a,b]$ satisfy $f(c)=0$. Let $g:[a,b]\to\mathbb R$ be any function with a zero $d\in[a,b]$, and suppose $$ |f(x)-g(x)|\leq\delta \qquad (x\in[a,b]) $$ for some $\delta\geq0$. Then $$ |c-d|\leq\frac{\delta}{m}. $$

Proof: Apply the residual error estimate from the Unique Root with a Residual Error Bound theorem to $f$ at $x=d$. Its hypotheses on the endpoint signs are not needed for this application if the zero $c$ is already known: the estimate follows from the Mean Value Theorem between $c$ and $d$, with the case $c=d$ immediate. Thus $m|c-d|\leq |f(d)|$. Since $g(d)=0$, the uniform bound gives $$ |f(d)|=|f(d)-g(d)|\leq\delta. $$ Combining the inequalities yields $m|c-d|\leq\delta$, and division by $m>0$ proves the result. This proof also covers a zero at either endpoint, because the Mean Value Theorem is applied on the interval between the two zero locations and its intermediate point lies in $(a,b)$ when the locations differ. $\square$

The stability theorem requires that $f$ have the derivative lower bound; it does not require differentiability of $g$. Nor does it require $g$ to be monotone. It does require an actual zero $d$ of $g$, and a uniform error bound on the whole interval that includes $d$. Without the latter, the comparison at the point $d$ may not be controlled.

Worked Example: Bounding the Movement of a Root

On $[0,1]$, let $f(x)=x^2+x-1$ and $g(x)=f(x)+1/50$. As above, $f'(x)=2x+1\geq1$ on $(0,1)$. The functions differ everywhere by exactly $1/50$, so $$ |f(x)-g(x)|=\frac{1}{50}\qquad (x\in[0,1]). $$ The endpoint values of $g$ are $g(0)=-49/50<0$ and $g(1)=51/50>0$. Also $g'=f'\geq1$, so the combined root theorem gives a unique zero $d\in(0,1)$ of $g$, as well as the unique zero $c$ of $f$. The stability theorem then yields $$ |c-d|\leq\frac{1/50}{1}=\frac{1}{50}. $$ Thus a uniform change of at most $1/50$ in the function shifts this root by at most $1/50$ under the stated slope condition.

Approximate Zeros and Convergent Computations

A numerical procedure may produce a sequence of test points rather than one final answer. The root estimate immediately turns residuals into location errors, so it can certify convergence without requiring the test points themselves to be monotone. This is a separate conclusion from existence and uniqueness: it says that any sequence whose function values approach zero must approach the unique root, provided the same positive derivative bound is available.

Corollary (Vanishing Residuals Force Convergence to the Root): Under the hypotheses of the Unique Root with a Residual Error Bound theorem, let $(x_n)$ be any sequence in $[a,b]$. If $f(x_n)\to0$, then $x_n\to c$, where $c$ is the unique zero of $f$.

Proof: The theorem gives, for every positive integer $n$, $$ |x_n-c|\leq\frac{|f(x_n)|}{m}. $$ Since $m>0$ is fixed and $f(x_n)\to0$, the right-hand side tends to zero. Hence $|x_n-c|\to0$, which is exactly $x_n\to c$. $\square$

Worked Example: A Residual Test for a Sequence of Approximations

For $f(x)=x^2+x-1$ on $[0,1]$, the root $c$ is unique and the derivative is at least $1$. Suppose a computation gives points $x_n\in[0,1]$ with $|x_n^2+x_n-1|\leq 1/n$ for every positive integer $n$. Applying the root estimate separately at each $x_n$ gives $$ |x_n-c|\leq |x_n^2+x_n-1|\leq\frac1n. $$ Because $1/n\to0$, it follows that $x_n\to c$. The estimate proves more than eventual closeness: it gives an explicit error bound at every index. For example, whenever $n\geq100$, the error is at most $1/100$.

Proof Checks and Common Pitfalls

The structure of this argument is reusable, but each theorem has its own hypotheses. Before invoking a result, check that the interval on which it is used has the required continuity and differentiability. In the root estimate, the interval is the one with endpoints $x$ and $c$, not necessarily all of $[a,b]$. Continuity on $[a,b]$ and differentiability on $(a,b)$ guarantee the Mean Value Theorem hypotheses on that smaller interval, including when one endpoint is $a$ or $b$.

Another pitfall is to confuse a small residual $|f(x)|$ with a small distance $|x-c|$ without a slope condition. A function can be very flat near a zero, so a small function value alone need not force a comparably small distance. Here the lower bound $f'\geq m>0$ supplies the missing quantitative control. If $m$ is small, the resulting error bound $|f(x)|/m$ is correspondingly weaker.

Finally, keep existence and uniqueness logically separate. The endpoint sign change and continuity give a zero by the Intermediate Value Theorem, but do not by themselves rule out multiple zeros. Strict increase rules out multiple zeros, but the derivative information alone does not guarantee that the function crosses zero on the chosen interval. Once both facts are established, the Mean Value Theorem adds a third conclusion: a practical estimate that can be used for approximation and stability.

Check Your Understanding

For each question, identify the exact hypothesis or theorem that supports the step.

Which assumptions give existence of a zero in the combined root theorem, and which give uniqueness?
In the residual estimate, why is the Mean Value Theorem applicable when the test point is an endpoint of $[a,b]$?
For $f(x)=x^2+x-1$ on $[0,1]$, what error bound follows from testing $x=3/5$?
In the stability theorem, why is it enough to know that $g(d)=0$ and that $f$ and $g$ are uniformly within $\delta$?
If $f(x_n)\to0$, which inequality proves that $x_n$ converges to the unique root?