Question

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)

Answer 1

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Solução!

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

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



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

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