Răspuns:
1. f(a)+3= f(5) f(x)=3x+11 se inlocuieste x cu a, 5
f(5)= 3*5+11 =15+11=26 f(a)= 3a+11
3a+11+3= 26 3a+14=26 3a= 26-14 =12
3a=12 => a= 12:3= 4 a=4
2. f(x)=5x+3 f(-4)-f(-3)- f(-2) +f(-1)= 0 se inlocuieste x cu -4, -3, -2, -1
f(-4)= 5* (-4)+3 = -20+3= -17 f(-3)= 5*(-3)+3= -15+3= -12
f(-2)= 5*(-2)+3= -10+3= -7 f(-1)= 5*(-1)+3= -5+3= -2
f(-4)-f(-3)- f(-2) +f(-1)= -17- (-12) - (-7) +(-2)=
-17+12+7-2= -19+19= 0
3. f(x)=6x+3 f([tex]\sqrt{2}[/tex])= 6*[tex]\sqrt{2}[/tex] + 3= 3( 2[tex]\sqrt{2}[/tex]+1)= 3*(2*1,41+1)= 3*(2.82+1)= 3*3.82= 11,46
f([tex]\sqrt{3}[/tex])= 6*[tex]\sqrt{3}[/tex] +3= 3(2 [tex]\sqrt{3}[/tex]+ 1)= 3* (2*1,73+1)= 3*(3,46+1)= 3*4,46 =13,38
f([tex]\sqrt{2}[/tex])< f([tex]\sqrt{3}[/tex])
4. f(x)=-6x-12 f(-1) = (-6)*(-1)-12= 6-12= -6
f(0)= -6*0-12= -12 f(-2) = (-6)*(-2)-12= 12-12= 0
coordonate sunt A(0, -12) B( -2, 0)
5. f(x)= 2x+2013 f(u)= 2u+2013 f(5)= 2*5+2013= 10+2013= 2023
[tex]\frac{f(u) -f(5)}{u-5}[/tex]= [tex]\frac{2u+2013 -2023}{u-5}[/tex]= [tex]\frac{2u-10}{u-5}[/tex]= [tex]\frac{2(u-5)}{u-5}[/tex]= 2
6. f(x)=x+3 2f(x)+f(2) ≥4f(1)
f(2)= 2+3=5 f(1)= 1+3=4
2x+6 +5 ≥4*4 2x+11≥ 16 2x≥16-11 2x≥5
x≥5/2 x∈[5/2, ∞)
