If g(x)=1+√[3]x then a function f such that f(g(x))=3-3(√[3]x)+x is

Question

If g(x)=1+√[3]x then a function f such that f(g(x))=3-3(√[3]x)+x is (A) f(x)=x3-3 x2+x+5 (B) f(x)=x3+3 x2-x-5 (C) f(x)=x3-3 x2+5 (D) f(x)=x3-3 x2+3 x+3

Tardigrade Admin · Accepted Answer

Let g(x)=1+√[3]x=y ⇒ x1 / 3=y-1 text so f(g(x))=3-3 x1 / 3+x ⇒ f(y)=3-3(y-1)+(y-1)3 =y3-3 y2+5

Q. If $g (x) = 1 + 3 x$ then a function $f$ such that $f (g (x)) = 3 - 3 (3 x) + x$ is

$f (x) = x^{3} - 3 x^{2} + x + 5$

$f (x) = x^{3} + 3 x^{2} - x - 5$

$f (x) = x^{3} - 3 x^{2} + 5$

$f (x) = x^{3} - 3 x^{2} + 3 x + 3$

Let $g (x) = 1 + 3 x = y \Rightarrow x^{1/3} = y - 1$
$so f (g (x)) = 3 - 3 x^{1/3} + x$
$\Rightarrow f (y) = 3 - 3 (y - 1) + (y - 1)^{3}$
$= y^{3} - 3 y^{2} + 5$

Q. If g(x)=1+3x​ then a function f such that f(g(x))=3−3(3x​)+x is

f(x)=x3−3x2+x+5

f(x)=x3+3x2−x−5

f(x)=x3−3x2+5

f(x)=x3−3x2+3x+3