Question

Let f(x) = 4x + 2 and g(x) = 2x2-4. Find the formula for the composition function g f.

Answer 1

Step-by-step explanation:

g(f(x)) => g(4x+2) let's take 4x+2 as y. So

$g(y) = 2 {y}^{2} - 4$

Substitute y=> 4x+2 in g(y) then we get

$g(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >$

$g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4$

$g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >$

$g(f(x)) = 32 {x}^{2} +32x + 4$

Hope it helps you.

Answer 2

Answer:

g(f(x)) =32x+32x+4

Step-by-step explanation:

g(f(x)) => g(4x+2) let's take 4x+2 as y. So

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we get

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2)

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >g(f(x))=32x

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >g(f(x))=32x 2

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >g(f(x))=32x 2 +8+32x−4=>

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >g(f(x))=32x 2 +8+32x−4=>g(f(x)) = 32 {x}^{2} +32x + 4g(f(x))=32x

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >g(f(x))=32x 2 +8+32x−4=>g(f(x)) = 32 {x}^{2} +32x + 4g(f(x))=32x 2

g(f(x)) => g(4x+2) let's take 4x+2 as y. Sog(y) = 2 {y}^{2} - 4g(y)=2y 2 −4Substitute y=> 4x+2 in g(y) then we getg(4x + 2) = 2 {(4x + 2)}^{2} - 4 = >g(4x+2)=2(4x+2) 2 −4=>g(f(x)) = 2(16 {x}^{2} + 4 + 2.2.4x) - 4g(f(x))=2(16x 2 +4+2.2.4x)−4g(f(x)) = 32 {x}^{2} + 8 + 32x - 4 = >g(f(x))=32x 2 +8+32x−4=>g(f(x)) = 32 {x}^{2} +32x + 4g(f(x))=32x 2 +32x+4

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