Resposta:

c)2.05.................

Resposta:

Alternativa a)

Explicação passo-a-passo:

Fazendo (x − 7)/(√x + √7) = 1, temos

(√x − √7)(√x + √7)/(√x + √7) = 1

(√x − √7)(√x + √7)/(√x + √7) = 1, [Eliminamos os termos semelhantes]

(√x − √7)/1 = 1

√x − √7 = 1

√x = 1 + √7

x = (1 + √7)^2

Substituindo (1 + √7)^2 em f, temos:

f[(x − 7)/(√x + √7)] = √(log x), [Já sabemos que f[(x − 7)/(√x + √7)] = f[√x − √7]

f[√x − √7] = √(log x)

f[√(1 + √7)^2) − √7] = √(log[(1 + √7)^2 ])

f[1 + √7 − √7] = √(2log[1 + √7])

f[1] = √(2log[1 + √7])

f[1] ≈ 1.059988, [Arredondando para 1.06]

f[1] ≈ 1.06

Sabendo f[√x − √7], repetimos o mesmo processo para cada caso, agora, para f(2):

√x − √7 = 2

√x = 2 + √7

x = (2 + √7)^2

f[√x − √7] = √(log x)

f[√(2 + √7)^2 − √7] = √(log[(2 + √7)^2 ])

f[2 + √7 − √7] = √(2log[2 + √7])

f[2] ≈ 1.15

Para f(3):

√x − √7 = 3

x = (3 + (√7))^2

f[√x − √7] = √(log x)

f[√(3 + √7)^2 − √7] = √(log[(3 + (√7))^2 ])

f[3 + √7 − √7] = √(2log[3 + √7])

f[3] = √(2log[3 + √7])

f[3] ≈ 1.23

Para f(4):

√x − √7 = 4

√x = 4 + √7

x = (4 + √7)^2

f[√x − √7] = √(log x)

f[√(4 + √7)^2 − √7] = √(log[(4 + √7)^2])

f[4 + √7 − √7] = √(2log[4 + √7])

f[4] = √(2log[4 + √7])

f[4] ≈ 1.28

Se f[(x − 7)/(√x + √7)] = √(log x), vamos determinar f[(f(1) + f(2))/(f(3) + f(4))] igualando:

(x − 7)/(√x + √7) = (f(1) + f(2))/(f(3) + f(4))

(x − 7)/(√x + √7) = (1.06 + 1.15)/(1.23 + 1.28)

(x − 7)/(√x + √7) = 0.88

√x − √7 = 0.88

x = (0.88 + √7)^2

f[(x − 7)/(√x + √7)] = √(log x)

f(√x − √7) = √(log[(0.88 + √7)^2])

f[(√(0.88 + √7)^2 − √7] = √(2log[0.88 + √7])

f[0.88 + √7 − √7] = √(2log[0.88 + √7])

f[0.88] = √(2log[0.88 + √7])

f(0.88) ≈ 1.05

∴ Pode-se dizer que, sendo f[(x − 7)/(√x + √7)] = √(log x), f[(f(1) + f(2)))/(f(3) + f(4)))] ≈ 1.05