Question

Considere f uma função, determine: $f(x)= 3x^{2} - x + 4\\a) f(1)\\b)f(-1)\\c)f(0)\\d)\frac{1}{2}$

Answer 1

Explicação passo-a-passo:

a)

f(1) = 3.1² - 1 + 4

f(1) = 3.1 - 1 + 4

f(1) = 3 - 1 + 4

f(1) = 6

b)

f(-1) = 3.(-1)² - (-1) + 4

f(-1) = 3.1 + 1 + 4

f(-1) = 3 + 1 + 4

f(-1) = 8

c)

f(0) = 3.0² - 0 + 4

f(0) = 0 - 0 + 4

f(0) = 4

d)

f(1/2) = 3.(1/2)² - 1/2 + 4

f(1/2) = 3.1/4 - 1/2 + 4

f(1/2) = 3/4 - 1/2 + 4

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

f(1/2) = 17/4

Answer 2

Oie, Td Bom?!

• Seja a função:

$f(x)= 3x^{2} - x + 4$

a)

$f(1) = 3 \: . \: 1 {}^{2} - 1 + 4$

$f(1) = 3 \: . \: 1 - 1 + 4$

$f(1) = 3 - 1 + 4$

$f(1) = 7 - 1$

$f(1) = 6$

b)

$f( - 1) = 3 \: . \: ( - 1) {}^{2} - ( - 1) + 4$

$f( - 1) = 3 \: . \: 1 + 1 + 4$

$f( - 1) = 3 + 1 + 4$

$f( - 1) = 4 + 4$

$f( - 1) = 8$

c)

$f(0) = 3 \: . \: 0 {}^{2} - 0 + 4$

$f(0) = 3 \: . \: 0 + 4$

$f(0) = 0 + 4$

$f(0) = 4$

d)

$f( \frac{1}{2} ) = 3 \: . \: ( \frac{1}{2} ) {}^{2} - \frac{1}{2} + 4$

$f( \frac{1}{2} ) = 3 \: . \: \frac{1}{4} - \frac{1}{2} + 4$

$f( \frac{1}{2} ) = \frac{3}{4} - \frac{1}{2} + 4$

$f( \frac{1}{2} ) = \frac{3 - 2 + 16}{4}$

$f( \frac{1}{2} ) = \frac{17}{4}$

Att. Makaveli1996