Question

uma função real é dada pela lei f(x)=x-3x+5 determine os valores de a, tais que:f(a+1)=1/3 f(2a)

Answer 1

$<var>f(x) = x^{2}-3x+5 f(a+1) = (a+1)^{2}-3(a+1)+5</var>$

 

$<var>f(a+1) = a^{2}+2a+1 -3a-3+5 f(a+1) = a^{2}-a+3</var>$

 

$<var>f(2a) = (2a)^{2}-3(2a)+5 => f(2a) = 4a^{2} - 6a+5 </var>$

 

$<var>1/3f(2a) = \frac{4a^{2}-6a+5}{3} f(a+1) = 1/3f(2a) a^{2}-a+3 = \frac{4a^{2}-6a+5}{3} 3a^{2}-3a+9 = 4a^{2} - 6a +5 a^{2}-3a-4=0 Delta = b^{2} - 4ac Delta = 9 +16 = 25 </var>$

 

$<var> a = \frac{3 +-\sqrt{25}}{2} ===> a' =\frac{3+5}{2} ===> a' = 4  a"=\frac{3-5}{2} ==> a"= -1</var>$