Matemática, perguntado por káh123linda, 1 ano atrás

seja f a função Z em Z defenida por f(x)=x^2-3x-2 calcule:
a)f(2),
b)f(-1)
c)raiz quadrada de 3,
d)(1/2)

Soluções para a tarefa

Respondido por Usuário anônimo
0
Bom dia Linda!

Solução!

A)~~\\\\
f(x)= x^{2} -3x-2\\\\\\
f(2)=(2)^{2}-3(2)-2\\\\\\\
 f(2)=4-6-2\\\\\\\
\boxed{f(2)=-4}\\\\\\\\
B)~~\\\\
f(x)= x^{2} -3x-2\\\\\\\\
f(-1)= (-1)^{2} -3(-1)-2\\\\\\\
f(-1)=1 +3-2\\\\\\\
f(-1)=4-2\\\\\\\
\boxed{f(-1)=2}\\\\\\\

c)~~\\\\\
f(x)= x^{2} -3x-2\\\\\\
f( \sqrt{3} )= ( \sqrt{3}) ^{2} -3( \sqrt{3}) -2\\\\\\
f( \sqrt{3} )= 3 -3( \sqrt{3}) -2\\\\\\
f( \sqrt{3} )= 3-2 -3( \sqrt{3}) \\\\\\
\boxed{f( \sqrt{3} )= 1 -3( \sqrt{3})} \\\\\\
Ou\\\\
f( \sqrt{3} )= 1 -3(1,73) \\\\\\
f( \sqrt{3} )= 1 -5,19 \\\\\\
\boxed{f( \sqrt{3} )= -4,19} \\\\\\



D)~~\\\\
f(x)= x^{2} -3x-2\\\\\\
f( \frac{1}{2} )= ( \frac{1}{2}) ^{2} -3( \frac{1}{2}) -2\\\\\\
f( \frac{1}{2} )= ( \frac{1}{4})  -( \frac{3}{2}) -2\\\\\\
f( \frac{1}{2} )= ( \frac{1-6-8}{4}) \\\\\\
f( \frac{1}{2} )= ( \frac{1-14}{4}) \\\\\\
\boxed{f( \frac{1}{2} )= -\frac{13}{4}} \\\\\\

Bom dia!
Bons estudos!


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