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If f(x) = 2x^2 – 3x + 1 and g(x) = x + 1, then
f(g(x)) =

Sagot :

[tex]\text{Given that,}~ f(x) =2x^2 -3x +1~ \text{and}~ g(x) = x+1[/tex]

[tex]f(g(x))\\\\=f(x+1)\\\\=2(x+1)^2 -3(x+1) +1\\\\=2(x^2 +2x +1) -3x -3 +1\\\\=2x^2 +4x +2-3x -2\\\\=2x^2 +x\\\\=x(2x+1)[/tex]

s1m1

Answer:

f(g(x)) = 2x² + x

Step-by-step explanation:

f(x) = 2x² – 3x + 1 and g(x) = x + 1

→substitute  g(x)  for  x + 1

f(g(x)) = f (x + 1)

→substitute x for x+1 into the f(x) equation

f(g(x)) = f (x + 1)  = 2(x+1)² – 3(x+1) + 1

→calculate

f(g(x)) =  2(x+1)² – 3(x+1) + 1 , square x+1

f(g(x)) = 2( x²+2x+1) -3(x+1) +1 , distribute in parenthesis

f(g(x)) = 2x²+4x+2 -3x -3 +1, combine like terms

f(g(x)) = 2x² + x

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