(C) Given,$f(x) = ax + b$ and $g(x) = cx^3 + d$. $(f \circ g)(x) = f(g(x))$. $(f \circ g)(x) = a(cx^3 + d) + b = acx^3 + ad + b$. Let $y = (f \circ g)

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$\left( \frac{x - ad - b}{ac} \right)^{\frac{1}{3}}$

(C) Given,$f(x) = ax + b$ and $g(x) = cx^3 + d$. $(f \circ g)(x) = f(g(x))$. $(f \circ g)(x) = a(cx^3 + d) + b = acx^3 + ad + b$. Let $y = (f \circ g)(x) = acx^3 + ad + b$. To find the inverse,solve for $x$ in terms of $y$: $y - ad - b = acx^3$. $x^3 = \frac{y - ad - b}{ac}$. $x = \left( \frac{y - ad - b}{ac} \right)^{\frac{1}{3}}$. Replacing $y$ with $x$,we get $(f \circ g)^{-1}(x) = \left( \frac{x - ad - b}{ac} \right)^{\frac{1}{3}}$.

Class 12 Mathematics Relation and Function Composition of Functions

Medium

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x) = ax + b$ $(a \neq 0)$ for all $x \in R$ and $g(x) = cx^3 + d$ $(c \neq 0)$ for all $x \in R$,then $(f \circ g)^{-1}(x) =$

AP EAMCET 2021 All AP EAMCET

A
$\left( \frac{x - ad + b}{ac} \right)^{\frac{1}{2}}$
B
$\left( \frac{x + ad - b}{ac} \right)^{\frac{1}{3}}$
C
$\left( \frac{x - ad - b}{ac} \right)^{\frac{1}{3}}$
D
$\left( \frac{x + ad + b}{ac} \right)^{\frac{1}{3}}$

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If $f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions defined by $f(x) = ax + b$ $(a \neq 0)$ for all $x \in R$ and $g(x) = cx^3 + d$ $(c \neq 0)$ for all $x \in R$,then $(f \circ g)^{-1}(x) =$

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