> #mathematics , #functions , #real-numbers > Let f, g and h be mapping on the set of real number defined by F (X) =2x+1 G(x) =x^2-3 and H(x) = 3x+2. Show that (FoG)=Fo (GoH).?
Anonymous
Mar 21

Let f, g and h be mapping on the set of real number defined by F (X) =2x+1 G(x) =x^2-3 and H(x) = 3x+2. Show that (FoG)=Fo (GoH).?

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ChatGPT
Mar 21
To show that (FoG) = Fo(GoH), we need to first find the compositions of F and G, and F and H, and then compare the results.

1. (FoG):
(FoG)(x) = F(G(x))
(FoG)(x) = F(x^2 - 3) [Substitute G(x) into F(x)]
(FoG)(x) = 2(x^2 - 3) + 1 [Substitute F(x) into the expression]
(FoG)(x) = 2x^2 - 6 + 1
(FoG)(x) = 2x^2 - 5

2. (GoH):
(GoH)(x) = G(H(x))
(GoH)(x) = G(3x + 2) [Substitute H(x) into G(x)]
(GoH)(x) = (3x + 2)^2 - 3 [Substitute G(x) into the expression]
(GoH)(x) = 9x^2 + 12x + 4 - 3
(GoH)(x) = 9x^2 + 12x + 1

Now, we need to find the composition of F and the result of (GoH):
F((GoH)(x)) = F(9x^2 + 12x + 1)
F((GoH)(x)) = 2(9x^2 + 12x + 1) + 1
F((GoH)(x)) = 18x^2 + 24x + 2 + 1
F((GoH)(x)) = 18x^2 + 24x + 3

Therefore, (FoG) = 2x^2 - 5 and Fo(GoH) = 18x^2 + 24x + 3. Since these two expressions are not equal, we can conclude that (FoG) is not equal to Fo(GoH).
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