Question

dau 50 puncte si coroana va rog ajutati-ma toate cele 3 puncte

Answer 1

1a)
f'(x)=(2√x-lnx)=(2√x)-(ln x)'=2/2√x-1/x=1/√x-1/x.
Amplificam cu √x prima fractie si obtinem:
f'(x)=(√x-1)/x

b)
tg: y-y0=m(x-x0)
y0=f(x0)=f(1)=2
m=f'(x0)=f'(1)=(√1-1)/1=0
tg: y-2=0(x-1)
tg: y-2=0

c)
f'(x)=0
√x-1=0, x≥0
√x=1 |²
x=1
Facem tabelul de semn al functiei:
x |0_____________________1____________________∞
f' |-------------------------------------0+++++++++++++++++++
f  |          descrecatoare           2             crescatoare
f'(1/4)=(√1/4-1)/1/4=(1/2-1)*4=-1/2*4=-4/2=-2 <0
f'(4)=(√4-1)/4=(2-1)/4=1/4 >0
f(1)=2√1-ln 1=2-0=2
x=1 punct de minim=>
f(x)≥f(1)
2√x-lnx≥2
2√x≥2+ln x