exercice dérivation 5

salut




bonne chance à tous

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salut mon ami elpizjuan ( v ya3ni racine )
1) dala f(x) 9abla ll icht9a9 3ala lmajal [0,pi/2[
donc ladayna
f'(x)=-sinx/2vcosx +1
f'(x)=1/2vcosx(-sinx+2vcosx)
docn nastantijk ana
f'(x)>0 <==> -sinx+2vcosx>0
f'(x)>0 <==> sinx<2vcosx
wa na3lam ana 0<x<pi/2
donc sinx>0
f'(x)>0 <==> sin²x<4cosx
f'(x)>0 <==> 1-cos²x<4cosx (sin²x=1-cos²x)
f'(x)>0 <==> cos²x+4cosx>1
f'(x)>0 <==> cos²x+4cosx+4>5
f'(x)>0 <==> (cosx+2)²>5
f'(x)>0 <==> cosx+2>v5
f'(x)>0 <==> cosx>v5-2
ladayna 0<v5-2<1
donc youjd 3onsore wahid k mina l majal [0,pi/2[
bihayte cosk=v5-2
wa bima ana dala cosx tana9sia 9at3an 3ala l majal [0,pi/2[
fa ina
f'(x)>00 <==> x<k
wa bi nafse tari9a nebayn ana
f'(x)=0 <==> x=k
et f'(x)<0 <==> x>k
idan f' tan3adim fi k wa terayre icharteha
wa bitali fa ina k hia mitraf dala f

2) mahma yakon x min [0.k[
f'(x)>0
waminh f'(x) tazaydia 3ala l majal [0,k[
idan likoli x min [0,k[ f(x)<=f(k)
likoli x min ]k,pi/2[ f'(x)<0
donc f' tana9osia 3ala l majal [k,pi/2[
donc likoli x min [k, pi/2[ f(x)<=f(k)
donc likoli x min [0,pi/2[ f(x)<=f(k)
ladayna f(k)=vcosk+k
=vv5-2k
wa bima ana k<pi/2 fa ina k<2 li ana pi/2<2
donc f(x)<vv5-2+2 li koli x min [0,pi/2[

bon chance

bravo 3issam

wliti m7ayeh

hhh

bonne chance !