Question Number 71816 by psyche last updated on 20/Oct/19
![suppose that f is continuous and differentiable in (a,b) if f′(x) =0 ,∀ x∈(a,b) then show that f is constant on [a,b].](https://www.tinkutara.com/question/Q71816.png)
$${suppose}\:{that}\:{f}\:{is}\:{continuous}\:{and}\:{differentiable}\:{in}\:\left({a},{b}\right)\:{if}\:{f}'\left({x}\right)\:=\mathrm{0}\:,\forall\:{x}\in\left({a},{b}\right)\:{then}\:{show}\:{that}\:{f}\:{is}\:{constant}\:{on}\:\left[{a},{b}\right]. \\ $$
Answered by mind is power last updated on 20/Oct/19
![assum f is not consrante ∃x,y suche that 1≥y≠x≥0 f(x)≠f(y) use mean value theorem for f in [x,y]⇒ ∃c∈[x,y] such that ((f(y)−f(x))/(y−x))=f′(c)⇒f(y)−f(x)=(y−x)f′(c) but f′(c)=0 cause c∈[0,1] ⇒f(y)−f(x)=0⇒f(y)=f(x) absurd⇒∀(x,y)∈[0,1]^2 f(x)=f(y) f constant](https://www.tinkutara.com/question/Q71830.png)
$$\mathrm{assum}\:\mathrm{f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{consrante} \\ $$$$\exists\mathrm{x},\mathrm{y}\:\mathrm{suche}\:\mathrm{that}\:\mathrm{1}\geqslant\mathrm{y}\neq\mathrm{x}\geqslant\mathrm{0}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\neq\mathrm{f}\left(\mathrm{y}\right) \\ $$$$\mathrm{use}\:\:\mathrm{mean}\:\mathrm{value}\:\mathrm{theorem}\:\mathrm{for}\:\mathrm{f}\:\mathrm{in}\:\left[\mathrm{x},\mathrm{y}\right]\Rightarrow \\ $$$$\exists\mathrm{c}\in\left[\mathrm{x},\mathrm{y}\right]\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{\mathrm{f}\left(\mathrm{y}\right)−\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{y}−\mathrm{x}}=\mathrm{f}'\left(\mathrm{c}\right)\Rightarrow\mathrm{f}\left(\mathrm{y}\right)−\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{y}−\mathrm{x}\right)\mathrm{f}'\left(\mathrm{c}\right) \\ $$$$\mathrm{but}\:\mathrm{f}'\left(\mathrm{c}\right)=\mathrm{0}\:\mathrm{cause}\:\mathrm{c}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{y}\right)−\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{absurd}\Rightarrow\forall\left(\mathrm{x},\mathrm{y}\right)\in\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} \:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right) \\ $$$$\mathrm{f}\:\mathrm{constant} \\ $$
Commented by psyche last updated on 24/Oct/19
$${thanks} \\ $$