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(B) Let $h(x) = (fog)(x)$.Given $h^{-1}(x) = \left(\frac{x-7}{2}\right)^{1/3}$.To find $h(x)$,let $y = \left(\frac{x-7}{2}\right)^{1/3}$. Then $y^3 =

Vedclass · Accepted Answer

$2039$

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(B) Let $h(x) = (fog)(x)$.Given $h^{-1}(x) = \left(\frac{x-7}{2}\right)^{1/3}$.To find $h(x)$,let $y = \left(\frac{x-7}{2}\right)^{1/3}$. Then $y^3 = \frac{x-7}{2}$,which implies $x = 2y^3 + 7$. Thus,$h(x) = 2x^3 + 7$.We have $(fog)(x) = f(g(x)) = a(x^b + c) - 3 = ax^b + ac - 3$.Comparing $ax^b + ac - 3 = 2x^3 + 7$,we get $a=2$,$b=3$,and $ac-3=7$,so $ac=10$. Since $a=2$,$c=5$.Now,$(fog)(ac) = (fog)(10) = 2(10)^3 + 7 = 2000 + 7 = 2007$.Next,$(gof)(x) = g(f(x)) = g(ax-3) = (ax-3)^b + c = (2x-3)^3 + 5$.Then $(gof)(b) = (gof)(3) = (2(3)-3)^3 + 5 = (3)^3 + 5 = 27 + 5 = 32$.Finally,$(fog)(ac) + (gof)(b) = 2007 + 32 = 2039$.

For some $a, b, c \in N$,let $f(x)=ax-3$ and $g(x)=x^b+c$,$x \in R$. If $(fog)^{-1}(x)=\left(\frac{x-7}{2}\right)^{1/3}$,then $(fog)(ac) + (gof)(b)$ is equal to $..........$

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