Question

Considere a função f(x)= X²+10 e determine:
$f(x) = {x}^{2} + 10$

A) f(2)
B) f(-7)
C) f(0)
D) f(6)
E) f(-3)​

Answer 1

Letra A:

f(2) = 2² + 10

f(2) = 4 + 10

f(2) = 14

Letra B:

f(- 7) = - 7² + 10

f(- 7) = 49 + 10

f(- 7) = 59

Letra C:

f(0) = 0² + 10

f(0) = 0 + 10

f(0) = 10

Letra D:

f(6) = 6² + 10

f(6) = 36 + 10

f(6) = 46

Letra E:

f(- 3) = - 3² + 10

f(- 3) = 9 + 10

f(- 3) = 19

Answer 2

$\Large\mathsf\displaystyle{}f\left(x\right) = {x}^{2} + 10$

Vamos lá

$\Large\mathsf\displaystyle{}a)f\left(2\right) = {2}^{2} + 10 \\ \Large\mathsf\displaystyle{}f\left(2\right) = 4 + 10 \\ \Large\mathsf\displaystyle{}f\left(2\right) = 14$

$\Large\mathsf\displaystyle{}b)f\left( - 7\right) =\left( - 7\right)^{2} + 10 \\ \Large\mathsf\displaystyle{}f\left( - 7\right) = 49 + 10 \\\Large\mathsf\displaystyle{}f \left( - 7\right) = 59$

$\Large\mathsf\displaystyle{}c)f\left(0\right) = {0}^{2} + 10 \\ \Large\mathsf\displaystyle{}f \left(0\right) = 0 + 10 \\ \Large\mathsf\displaystyle{}f\left(0\right) = 10$

$\Large\mathsf\displaystyle{}d)f\left(6\right) = {6}^{2} + 10 \\\Large\mathsf\displaystyle{}f\left(6\right) = 36 + 10 \\ \Large\mathsf\displaystyle{}f\left(6\right) = 46$

$\Large\mathsf\displaystyle{}e)f\left( - 3\right) =\left( - 3\right) ^{2} + 10 \\ \Large\mathsf\displaystyle{}f\left( - 3\right) = 9 + 10 \\ \Large\mathsf\displaystyle{}f\left( - 3\right) = 19$

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