Dada a função F(x)=3x²-5x, determine o valor de f(4)+ 2 f(0) -f(-3) f(0)+5.f(-1)+f(4) f(-3) -2.f(-1) + f(1)
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f(-3)=3×-3²-5×-3=27+15=42
f(-1)=3×-1²-5×-1=8
f(0)=3×0²-5×0=0
f(1)=3×1²-5×1= -2
f(2)=3×2²-5×2=2
f(3)=3×3²-5×3=12
f(4)=3×4²-5×4=28
f(5)=3×5²-5×5=50
28+0+0-42+0+40+28+42-16-2=78
f(-1)=3×-1²-5×-1=8
f(0)=3×0²-5×0=0
f(1)=3×1²-5×1= -2
f(2)=3×2²-5×2=2
f(3)=3×3²-5×3=12
f(4)=3×4²-5×4=28
f(5)=3×5²-5×5=50
28+0+0-42+0+40+28+42-16-2=78
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