Dada a função: f=(x)=x^2-3x-4
A) f(√3/5)
B)f (-1/2)+f (3)
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Vamos substituir e resolver.
A) f(√3/5) = ( √3/5)² -3.(√3/5) - 4
f(√3/5) = 3/5 -3√3/5 - 4
f(√3/5) = -17/5 - 3√3/5
B) f(-1/2) + f(3)= (1/2)²+3.1/2-4 + 3²-3.3-4
f(-1/2) + f(3) = 1/4+3/2 -4+9-9-4
f(-1/2) + f(3) = 14/8 -8
f(-1/2) + f(3) = 50/8
f(-1/2) + f(3) = 25/4
A) f(√3/5) = ( √3/5)² -3.(√3/5) - 4
f(√3/5) = 3/5 -3√3/5 - 4
f(√3/5) = -17/5 - 3√3/5
B) f(-1/2) + f(3)= (1/2)²+3.1/2-4 + 3²-3.3-4
f(-1/2) + f(3) = 1/4+3/2 -4+9-9-4
f(-1/2) + f(3) = 14/8 -8
f(-1/2) + f(3) = 50/8
f(-1/2) + f(3) = 25/4
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