Given that $f (x) = x - 1$ $f(x)=x-1$ and $(g \circ f) (x) = 3 x^{2} + 2$ $(g\circ f)(x)=3x^2 + 2$ , find the function of $g$ $g$ in similar form? Thank you

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Answer 1 · 2018-08-11T16:50:42

One possibility is $g (x) = 3 x^{2} + 6 x + 5$ $g(x) = 3x^2+6x+5$

There are infinite $g$ $g$ 's which satisfy this condition. Let us consider as $g$ $g$ pertaining to the polynomials, for instance let

$g (x) = a x^{2} + b x + c$ $g(x) = a x^2+b x+c$

then

$g (f (x)) = a {(f (x))}^{2} + b f (x) + c = a {(x - 1)}^{2} + b (x - 1) + c$ $g(f(x)) = a (f(x))^2+b f(x) + c = a(x-1)^2+b(x-1)+c$

so

$g (f (x)) = (a - 3) x^{2} + (b - 2 a) x + a + c - b + 2 = 3 x^{2} + 2$ $g(f(x)) = (a-3)x^2+(b-2a)x+a+c-b+2 = 3x^2+2$

and now comparing coeficients

$a = 3, b = 6, c = 5$ $a = 3, b = 6, c = 5$

hence $g (x) = 3 x^{2} + 6 x + 5$ $g(x) = 3x^2+6x+5$

Given that f(x)=x−1f(x)=x-1 and (g∘f)(x)=3x2+2(g\circ f)(x)=3x^2 + 2, find the function of gg in similar form? Thank you