Question

Considere f uma função de R em R definida pela lei f(x) = 3x²-x+4. Calcule:

a) f(1)
b) f(-1)
c) f(0)
d) f(1/2)
e) f(raiz quadrada de 2)

Answer 1

a) f(1) = 3(1)² -1+4
             3 - 1 + 4 = 6

b) f(-1) = 3(-1)² - (-1) + 4
                3 + 1 + 4= 8

c)f(0) = 3(0)² - 0 + 4 
             0 - 0 + 4 = 4

d) f(1/2) = 3(1/2)² - (1/2) + 4
                  3(1/4)- 1/2 + 4 
                     3/4 - 1/2 +  4
                   3/4 - 2/4 + 4
                        1/4 + 4 
                          1/4 + 16/4= 17/4

Answer 2

Olá,

na função f(x)=3x² - x + 4

f(1)

$f(1)=3\cdot1^2-1+4\\ f(1)=3\cdot1+3\\ f(1)=3+3\\\\ \huge\boxed{f(1)=6}$


f(-1)

$f(-1)=3\cdot(-1)^2-(-1)+4\\ f(-1)=3\cdot1+1+4\\ f(-1)=3+5\\\\ \huge\boxed{f(-1)=8}$


f(0)

$f(0)=3\cdot0^2-0+4\\ f(0)=3\cdot0+4\\ f(0)=0+4\\\\ \huge\boxed{f(0)=4}$


f(1/2)

$f\left( \dfrac{1}{2}\right)=3\cdot\left( \dfrac{1}{2}\right)^2- \dfrac{1}{2}+4\\\\ f\left( \dfrac{1}{2}\right)=3\cdot \dfrac{1}{4}- \dfrac{1}{2}+4\\\\ f\left( \dfrac{1}{2}\right)= \dfrac{3}{4}- \dfrac{1}{2}+ \dfrac{8}{2}\\\\ f\left( \dfrac{1}{2}\right)= \dfrac{3}{4}+ \dfrac{7}{2}\\\\ f\left( \dfrac{1}{2}\right)= \dfrac{3}{4}+ \dfrac{14}{4}\\\\\\ \huge\boxed{f\left( \dfrac{1}{2}\right)= \dfrac{17}{4}}$

f(√2)

$f( \sqrt{2} )=3\cdot( \sqrt{2})^2- \sqrt{2}+4\\ f( \sqrt{2} )=3\cdot2- \sqrt{2}+4\\ f( \sqrt{2} )=6+4- \sqrt{2}\\\\ \huge\boxed{f( \sqrt{2} )=10- \sqrt{2}}$


Tenha ótimos estudos ;D