Question

Galeraaa, Dada a função f(x)=x^2-x+4, calcule:

a) f(2)
b) f(0)
c) f(1/2)
d) f(√3)
e) f(-3)/(√2)

HELP ME, PLEASE.

Answer 1

a)

$f_{}(2) = {2}^{2} - 2 + 4 \\ f_{}(2) = 4 - 2 + 4 \\ f_{}(2) = 2 + 4 \\ f_{}(2) = 6$

b)

$f_{}(0) = {0}^{2} - 0 + 4 \\ f_{}(0) = 4$

c)

$f_{}( \frac{1}{2} ) = ( \frac{1}{2} {)}^{2} - \frac{1}{2} + 4 \\ \\ f_{}( \frac{1}{2} ) = \frac{1}{4} - \frac{1}{2} + 4 \\ \\ f_{}( \frac{1}{2} ) = \frac{ 1 - 2 + 16}{4} \\ \\ f_{}( \frac{1}{2} ) = \frac{15}{4}$

d)

$f_{}( \sqrt{3} ) = (\sqrt{3} {)}^{2} - \sqrt{3} + 4 \\ f_{}( \sqrt{3} ) = 3 - \sqrt{3} + 4 \\ f_{}( \sqrt{3} ) = 7 - \sqrt{3}$

e)

$f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = ( \frac{ - 3}{ \sqrt{2}^{} {}^{} } {)}^{2} - \frac{ - 3}{ \sqrt{2} } + 4 \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = (\frac{ - 3}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } {)}^{2} - \frac{ - 3}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } + 4 \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = ( \frac{ - 3 \sqrt{2} }{2} {)}^{2} - \frac{ - 3}{ 2} + 4 \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = \frac{9 \times 2}{4} - \frac{ - 3 \sqrt{2} }{2} + 4 \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = \frac{9}{2} - \frac{ - 3 \sqrt{2} }{2} + 4 \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = \frac{9 + 8}{2} - \frac{ - 3\sqrt{2} }{2} \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = \frac{17}{2} + \frac{3 \sqrt{2} }{2} \\ \\ f_{}( { \frac{ - 3}{ \sqrt{2} } } ) = \frac{17 + 3 \sqrt{2} }{2}$