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f-is-derivable-in-R-1-Demonstrate-that-if-f-is-pair-f-is-odd-unpair-1-Demonstrate-that-if-f-is-unpair-f-is-pair-Please-help-me-sirs-

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Question Number 79876 by mathocean1 last updated on 28/Jan/20
f is derivable in R. 1) Demonstrate that if f is pair , f ′ is odd(unpair). 1) Demonstrate that if f is unpair , f ′ is pair. Please help me sirs
fisderivableinR.1)Demonstratethatiffispair,f′isodd(unpair).1)Demonstratethatiffisunpair,f′ispair.Pleasehelpmesirs
Commented by MJS last updated on 29/Jan/20
if “pair” means f(−x)=f(x) and “unpair” means f(−x)=−f(x) then i.e. y=cos x and y=x^2 ; y=x^4 are pair y=sin x and y=x; y=x^3 are unpair draw them and try to understand how the slopes must look like
if“pair″meansf(−x)=f(x)and“unpair″meansf(−x)=−f(x)theni.e.y=cosxandy=x2;y=x4arepairy=sinxandy=x;y=x3areunpairdrawthemandtrytounderstandhowtheslopesmustlooklike
Answered by Henri Boucatchou last updated on 29/Jan/20
f is derivable for all x_0 ∈R, that′s lim_(x→x_0 ) ((f(x)−f(x_0 ))/(x−x_0 ))=f ′(x_0 ) f pair ⇒ lim_(x−→x_0 ) −((f(−x)−f(−x_0 ))/((−x)−(−x_0 )))=−f ′(−x_0 )=lim_(x→x_0 ) ((f(x)−f(x_0 ))/(x−x_0 ))=f ′(x_0 ) ⇒ f ′(−x_0 )=−f ′(x_0 ) e.i. f ′ is odd; f unpair ⇒ lim_(x→x_0 ) −((f(−x)−f(−x_0 ))/((−x)−(−x_0 )))=−f ′(−x_0 )=lim_(x→x_0 ) ((−[f(x)−f(x_0 )])/(x−x_0 ))=−f ′(x_0 ) ⇒−f ′(−x_0 )=−f ′(x_0 ) ⇒ f ′(−x_0 )=f ′(x_0 ) e.i. f ′ is pair.
fisderivableforallx0∈R,that′slimx→x0f(x)−f(x0)x−x0=f′(x0)fpair⇒limx−→x0−f(−x)−f(−x0)(−x)−(−x0)=−f′(−x0)=limx→x0f(x)−f(x0)x−x0=f′(x0)⇒f′(−x0)=−f′(x0)e.i.f′isodd;funpair⇒limx→x0−f(−x)−f(−x0)(−x)−(−x0)=−f′(−x0)=limx→x0−[f(x)−f(x0)]x−x0=−f′(x0)⇒−f′(−x0)=−f′(x0)⇒f′(−x0)=f′(x0)e.i.f′ispair.

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