34. Se considera functia f:R--->R, f(x)=x²-11x+30.Sa se calculeze f(0)·f(1)·...·f(6).

2
pai si de unde stiu eu care are 0?
Din f(x)=0=>x²-11x+30=0=>x=5 sau x=6=>f(5)=0 si f(6)=0.
Exemplu:f(x)= x²-1.Sa se calculeze f(0)·f(1)·...·f(2014). Din x²-1=0=> x=1 sau -1 adica f(1)=0 si f(-1)=0. In concluzie, f(0)·f(1)·...·f(2014)=0 deoarece unul din termeni adica f(1) este 0.
eu am f(0)= 0-0+30=30....nu da 0
am inteles

Răspunsuri

2014-08-10T14:02:44+03:00
 f:R--->R, f(x)=x²-11x+30
 f(0)·f(1)·...·f(6)=?
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f(0)=0²-11·0+30=30
f(1)=1²-11·1+30=20
f(2)=2²-11·2+30=12
f(3)=3²-11·3+30=6
f(4)=4²-11·4+30=2
f(5)=5²-11·5+30=0
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f(0)·f(1)......f(6)= 30·20·12·6·2·0 = 0 ( răspunsul final )
orice număr înmulțit cu zero da 0 de asta n-am inmultit la ultima
stiu
2014-08-10T14:23:27+03:00
 Avem functia  f:R--->R, f(x)=x²-11x+30
 f(0)·f(1)·...·f(6)=?
R/e

f(0)=0²-11*0+30=30
f(1)=1²-11*1+30=20
f(2)=2²-11*2+30=12
f(3)=3²-11*3+30=6
f(4)=4²-11*4+30=2
f(5)=5²-11*5+30=0
f(6)=6²-11*6+30=0

f(0)*......*f(6)= 30·*20*12* 6*2·0*0 = 0 
R/s 0