If f:R→R, g:R→R defined by f(x)=2x−3,g(x)=x3+5. Then (fog)−1(x)=

(fog)(x)=f[x3+5]=2(x3+5)−3=2x3+7 2x3+7=y⇒x3=y−72⇒x=(y−72)13 ∴(fog)−1(x)=(x−72)13

If f:R→R, g:R→R defined by f(x)=2x−3,g(x)=x3+5. Then (fog)−1(x)=

(fog)(x)=f[x3+5]=2(x3+5)−3=2x3+7 2x3+7=y⇒x3=y−72⇒x=(y−72)13 ∴(fog)−1(x)=(x−72)13

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If $f : R \to R, g : R \to R$ defined by $f (x) = 2 x - 3, g (x) = x^{3} + 5.$ Then ${(f o g)}^{- 1} (x) =$

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${(\frac{(x + 7)}{2})}^{\frac{1}{3}}$

${(x - \frac{7}{2})}^{\frac{1}{3}}$

${(\frac{x - 2}{7})}^{\frac{1}{3}}$

${(\frac{x - 7}{2})}^{\frac{1}{3}}$

answer is D.

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Detailed Solution

$(f o g) (x) = f [x^{3} + 5] = 2 (x^{3} + 5) - 3 = 2 x^{3} + 7$

$2 x^{3} + 7 = y \Rightarrow x^{3} = \frac{y - 7}{2} \Rightarrow x = {(\frac{y - 7}{2})}^{\frac{1}{3}}$

$∴ {(f o g)}^{- 1} (x) = {(\frac{x - 7}{2})}^{\frac{1}{3}}$

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If f:R→R, g:R→R defined by f(x)=2x−3,g(x)=x3+5. Then (fog)−1(x)=

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If $f : R \to R, g : R \to R$ defined by $f (x) = 2 x - 3, g (x) = x^{3} + 5.$ Then ${(f o g)}^{- 1} (x) =$