Question

(UCS) Uma função real do 1° grau é tal que F(0)=1+f(1) e f(-1)=2 -f(0). Então f(3) é igual a
a)-3
b)-5/2
c)-1
d)0
e)27

Answer 1

$\boxed{f(x)=ax+b}$
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$f(0)=1+f(1)\\a(0)+b=1+a(1)+b\\b=1+a+b\\0=1+a\\-1=a$

$f(x)=ax+b~~~\therefore~~~f(x)=(-1)x+b~~~\therefore~~~\boxed{f(x)=-x+b}$

Como f(-1) = 2 - f(0):

$f(-1)=2-f(0)\\(-1)(-1)+b=2-[(-1).0+b]\\1+b=2-b\\b+b=2-1\\2b=1\\b=1/2$

$\boxed{\boxed{f(x)=-x+\frac{1}{2}}}$
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Achando f(3):

$f(3)=-3+\dfrac{1}{2}\\\\\\f(3)=\dfrac{-6+1}{2}\\\\\\\boxed{\boxed{f(3)=-\dfrac{5}{2}}}$

Letra B