Question

dado f (x) = 3x+2 e g (x)=x²+1 pede-se

a) f(-2)

b) g(5)
c) f(0)+g(-3)
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     g(2)
d) f(g(-1)
e) g (g(-2)

Answer 1

d)
f [g(1)]

f(x) = 3x + 2
g(x) = x² + 1

f(g(x) = 3(x² + 1) + 2
f(g(-1)) = 3[(-1)² + 1] + 2
f(g(-1)) = 3.[1 + 1] + 2
f[g(-1)] = 3.2 + 2 = 6 + 2 = 8

R.: f[g(-1)] = 8
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e)
g(x) = x² + 1
g[g(x)] = (x²+1)² + 1
g[g(x)] = x⁴ + 2x² + 1 + 1

g[g(-2)] = (-2)⁴ + 2(-2)² + 2
g[g(-2)] = 16 + 2.4 + 2 = 16 + 8 + 2 = 24 + 2 = 26

R.: g[g( - 2)] = 26

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a)
f(x) = 3x + 2

x = - 2

f( - 2) = 3.(-2) + 2
f( - 2) = - 6 + 2

R.: f (- 2) = - 4
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b)
g(x) = x² + 1

g(5) = 5² + 1
g(5) = 25 + 1

R.: g(5) = 26
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c)
f(0) + g (-3)   =  2 + 10  =   12
   g(2)                  5            5

Resp.: 12/5

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f(x) = 3x + 2
f(0) = 3.0 + 2
f(0) = 0 + 2
f(0) = 2

g(x) = x² + 1
g(-3) = (-3)² + 1
g(-3) = 9 + 1
g(-3) = 10

g(2) = 2² + 1
g(2) = 4 + 1
g(2) = 5
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