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Rolle theorem states that if $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there is $c\in (a,b)$ with $f^{\prime }(c)=0$.

If $f$ is constant, any interior point works. Otherwise the extreme value theorem gives an attained maximum and minimum on the interval, and at least one of them differs from the common endpoint value. That extremum must occur at an interior point $c$. At an interior maximum, for small positive $h$ the difference quotient $[f(c+h)-f(c)]/h$ is nonpositive, while for negative $h$ it is nonnegative. Existence of the common derivative forces it to be both nonpositive and nonnegative, hence zero. The same argument with inequalities reversed applies to an interior minimum. This proves Rolle theorem, including the derivative-at-an-extremum step rather than assuming it.

Now let the nonconstant polynomial $p$ have distinct real roots $r_1<\cdots <r_k$, with multiplicities $m_1,\dots ,m_k$ adding to its degree $n$. If $p(x)=(x-r_j{)}^{m_j}{q}_j(x)$ and $q_j(r_j)\ne 0$, differentiation givesThe bracket is nonzero at $r_j$, so its multiplicity as a root of $p^{\prime }$ is exactly $m_j-1$. These roots already contribute ${\sum }_j(m_j-1)=n-k$ real roots counted with multiplicity. Rolle theorem supplies an additional root of $p^{\prime }$ in each of the $k-1$ disjoint intervals $(r_j,r_{j+1})$. These additional roots are distinct from each other and from all the $r_j$.

We have therefore exhibited at least $(n-k)+(k-1)=n-1$ real roots counted with multiplicity. The derivative has degree $n-1$, so there is no room for any additional nonreal roots. Every root of $p^{\prime }$ is real. This is the Rolle root count with multiplicities; for $n=1$ the nonzero constant derivative has no roots, as expected.

For failure of the converse, takeThe derivative has only the real root zero, of multiplicity two. But $p(x)=(x+1)(x^2-x+1)$ and the quadratic factor has roots $(1\pm i\sqrt{3})/2$, which are nonreal. Thus real-rootedness of the derivative does not imply real-rootedness of the original polynomial.

The mean value theorem states that a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$, with $a<b$, has a point $\xi \in (a,b)$ such thatFor the proof, subtract the secant line and put $h(t)=f(t)-f(a)-(t-a)[f(b)-f(a)]/(b-a)$. Then $h(a)=h(b)=0$. If $h$ is identically zero, every interior point works. Otherwise it has a positive maximum or a negative minimum. The extreme value theorem supplies that extremum, and its nonzero value makes its location interior. At an interior differentiable extremum, the difference quotients on the two sides have opposite weak signs, so their common limit is zero. Hence $h^{\prime }(\xi )=0$, proving the formula. This also gives the Rolle theorem argument rather than merely citing it.

One precise version of the Taylor theorem with Lagrange remainder is as follows. Let $n\ge 0$, let $f$ and its derivatives through order $n$ be continuous on the closed interval between distinct points $a$ and $x$, and suppose $f^{(n+1)}$ exists throughout its interior. Then some $\xi$ strictly between $a$ and $x$ satisfiesHere the endpoint derivatives are understood as restrictions of derivatives defined near the interval, or as their continuous one-sided extensions. No continuity of $f^{(n+1)}$ is required. For $x=a$ the equality is immediate with zero remainder.

To prove the formula, write the Taylor polynomial as $P_n(t)={\sum }_{k=0}^n{f}^{(k)}(a)(t-a{)}^k/k!$ and defineThen $h(a)=h(x)=0$, and $h^{(j)}(a)=0$ for $j=1,\dots ,n$. The Rolle theorem gives a zero ${\xi }_1$ of $h^{\prime }$ strictly between $a$ and $x$. If $n\ge 1$, apply the Rolle theorem to $h^{\prime }$ between $a$ and ${\xi }_1$, using $h^{\prime }(a)=0$, to get a zero ${\xi }_2$ of $h^{\prime \prime }$. Repeating produces a point ${\xi }_{n+1}$ strictly between $a$ and $x$ with $h^{(n+1)}({\xi }_{n+1})=0$. Each application is justified by continuity on the relevant closed interval and differentiability in its interior; the same construction works when $x<a$.

Since $P_n$ has degree at most $n$,Substituting this value of $K$ into its definition proves the displayed Taylor theorem with Lagrange remainder.

The Rolle theorem states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots <r_k$. The theorem gives a root of $f^{\prime }$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:

We prove the requested bound for positive roots of a polynomial by mathematical induction on the number of nonzero terms. In fact, the proof also works with roots counted according to their multiplicity. Write $V(p)$ for the number of sign changes and $N_{+}(p)$ for the positive-root count.

A one-term polynomial $a_1{x}^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:For $x>0$ this factor is nonzero, so $N_{+}(q)=N_{+}(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives

First suppose $a_1{a}_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q^{\prime })+1$. If $q$ has positive roots $r_1<\cdots <r_k$ of multiplicities $m_1,\dots ,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the Rolle theorem, an additional root between each pair. Thus the Rolle root count with multiplicities giveswhen there is a positive root. ConsequentlyIf there is no positive root, the desired bound is automatic.

Now suppose $a_1{a}_2>0$, so $V(q)=V(q^{\prime })$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,since $q(0)=a_1$ and the lowest-power term of $q^{\prime }$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The mean value theorem on $[u,r_1]$ gives a point $v\in (u,r_1)$ withBut $q^{\prime }(u)>0$. The intermediate value theorem applied to the polynomial $q^{\prime }$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. HenceBoth sign cases complete the induction:An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root. This is the upper-bound part of the Descartes' rule of signs. The multiplicity claim used above follows by differentiating $q(x)=(x-r{)}^{m}h(x)$ with $h(r)\ne 0$: the derivative has a factor $(x-r{)}^{m-1}$ and its remaining factor is nonzero at $r$.

The Lagrange cardinal polynomials areThus the Lagrange interpolation polynomial isIt has degree at most $n$ and takes the required values. Uniqueness follows because the difference of two such polynomials has $n+1$ distinct roots and degree at most $n$, so is zero. The source's phrase “degree $n$” must allow a smaller degree, for example when $f$ is constant.

Define the divided difference recursively by $f[x_i]=f_i$ andInduction on this recursion, or comparison of the leading coefficient in the Newton interpolation polynomial, givesFor completeness the induction works term by term: an interior term gets the coefficient $(1/(x_i-x_n)-1/(x_i-x_0))/(x_n-x_0)=1/((x_i-x_n)(x_i-x_0))$, multiplying the product over the other interior nodes; the two endpoint terms give the same formula. This is also the leading coefficient of $p$ directly from the cardinal formula, so $p^{(n)}=n!f[x_0,\dots ,x_n]$.

The function $f-p\in C^n[x_0,x_n]$ has $n+1$ distinct zeros. Repeated application of Rolle theorem gives $\xi \in (x_0,x_n)$ with $(f-p{)}^{(n)}(\xi )=0$ when $n\ge 1$. HenceFor $n=0$, this is simply $f[x_0]=f(x_0)$ and we take $\xi =x_0$.

Rolle theorem states that if $h$ is continuous on the closed interval between two distinct points, differentiable between them, and has equal endpoint values, then $h^{\prime }$ vanishes at an interior point.

Let $P_N(t)={\sum }_{j=0}^N{f}^{(j)}(0)t^j/j!$ and, for $x\ne 0$, putThen $h(x)=0$ and $h^{(j)}(0)=0$ for $0\le j\le N$. Applying Rolle's theorem successively $N+1$ times gives a point $\xi =\theta x$, $0<\theta <1$, where $h^{(N+1)}(\xi )=0$. Hencewhich is the stated Taylor theorem with Lagrange remainder.

If $f^{\prime }=0$, the $N=0$ case gives $f(x)=f(0)+x{f}^{\prime }(\theta x)=f(0)$. Thus $f$ is constant; equivalently this is the immediate corollary of the mean value theorem.

The Taylor theorem with Lagrange remainder states that if $f$ is $n$ times differentiable between $a$ and $x$, then some $\xi$ between $a$ and $x$ satisfies

Let $T_{n-1}$ denote the displayed polynomial and choose $K$ so thatsatisfies $F(x)=0$. By construction,Starting with the two zeros $a,x$ and applying Rolle theorem repeatedly, there is a $\xi$ between them with $F^{(n)}(\xi )=0$. Sincewe have $K=f^{(n)}(\xi )/n!$, proving the theorem.

Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

SetIts endpoint values are equal. Rolle theorem therefore supplies $\xi \in (a,b)$ withIf $g^{\prime }(\xi )=0$, then (1) and $g(b)\ne g(a)$ force $f^{\prime }(\xi )=0$, contrary to the hypothesis. Hence $g^{\prime }(\xi )\ne 0$, and division in (1) gives

The condition is necessary. On $[0,2\pi ]$, letHere $g(2\pi )-g(0)=\pi /2$ while $f(2\pi )-f(0)=0$, so the endpoint ratio is zero. ButWhere $g^{\prime }(x)\ne 0$, their ratio is $1/(1+\frac{1}{2}\cos x)$ and is never zero. At the remaining points both derivatives vanish, so the derivative ratio is undefined. Thus the conclusion fails when simultaneous zeros are allowed.

For any $n\ge 0$, take $f_n=T_{n+1}$, the Chebyshev polynomial of degree $n+1$. It has $n+2$ alternating extrema of magnitude one on $[-1,1]$. The Chebyshev alternation theorem shows that the zero polynomial is best from ${\mathcal{P}}_n$, with error one. Since it also lies in ${\mathcal{P}}_{n-1}$ when $n\ge 1$,

Now suppose $f^{(n)}(x)>0$ throughout $[-1,1]$ and, contrary to the claim, $E_{n-1}(f)=E_n(f)$. A best $p\in {\mathcal{P}}_{n-1}$ would then also be best in ${\mathcal{P}}_n$. Its error $e=f-p$ would have $n+2$ alternating extrema by the alternation theorem, hence at least $n+1$ distinct zeros. Applying the Rolle theorem $n$ times gives a zero ofcontradicting positivity. Therefore

If a nonzero real polynomial has $N$ roots in an open real interval, counted with multiplicity of a root, its derivative has at least $N-1$ roots in that interval, also counted with multiplicity, provided $N\ge 1$. Each root of multiplicity $m$ becomes a root of the derivative of multiplicity $m-1$. The Rolle theorem supplies one further root between each pair of distinct consecutive roots. Thus $k$ distinct roots of multiplicities $m_j$ give at least ${\sum }_j(m_j-1)+(k-1)=N-1$ derivative roots.