Rolle's Theorem

Rolle's Theorem. Suppose following three condition hold for function y=f(x)y={f{{\left({x}\right)}}}:

  1. function is defined and continuous on closed interval [a,b]{\left[{a},{b}\right]};
  2. exists finite derivative f(x){f{'}}{\left({x}\right)} on interval (a,b){\left({a},{b}\right)};
  3. f(a)=f(b){f{{\left({a}\right)}}}={f{{\left({b}\right)}}}.

then there exists point c{c} (a<c<b)\left({a}<{c}<{b}\right) such that f(c)=0{f{'}}{\left({c}\right)}={0}.

rolle theoremGeometrically Rolle's theorem means the following: if f(a)=f(b){f{{\left({a}\right)}}}={f{{\left({b}\right)}}} then there exists point c{c} at which tangent line is horizontal.

Note, that all three conditions are needed.

For function f(x)=x[x]{f{{\left({x}\right)}}}={x}-{\left[{x}\right]} on interval [0,1]{\left[{0},{1}\right]} first condition doesn't hold, because it is not continuous at x=1{x}={1}. And f(x)=1{f{'}}{\left({x}\right)}={1} on (0,1){\left({0},{1}\right)}, so there are no point c{c} from (0,1){\left({0},{1}\right)} such that f(c)=0{f{'}}{\left({c}\right)}={0}.

For function f(x)={xif0x121xif12x1{f{{\left({x}\right)}}}={\left\{\begin{array}{c}{x}{\quad\text{if}\quad}{0}\le{x}\le\frac{{1}}{{2}}\\{1}-{x}{\quad\text{if}\quad}\frac{{1}}{{2}}\le{x}\le{1}\\ \end{array}\right.} second condition doesn't hold, because derivative doesn't exist at x=12{x}=\frac{{1}}{{2}}. Aso f(x)=1{f{'}}{\left({x}\right)}={1} on (0,12){\left({0},\frac{{1}}{{2}}\right)} and f(x)=1{f{'}}{\left({x}\right)}=-{1} on (12,1){\left(\frac{{1}}{{2}},{1}\right)}, so there are no point c{c} from (0,1){\left({0},{1}\right)} such that f(c)=0{f{'}}{\left({c}\right)}={0}.

For function f(x)=x{f{{\left({x}\right)}}}={x} on interval [0,1]{\left[{0},{1}\right]} third condition doesn't hold, f(0)f(1){f{{\left({0}\right)}}}\ne{f{{\left({1}\right)}}}. And f(x)=1{f{'}}{\left({x}\right)}={1} everywhere, so there are no point c{c} from (0,1){\left({0},{1}\right)} such that f(c)=0{f{'}}{\left({c}\right)}={0}.