Let f: R arrow R , g: R arrow R be two functions such that f(x)=2 x-3, g(x)=x3+5. The function ( text fog )-1(x) is equal to

Question

Let f: R arrow R , g: R arrow R be two functions such that f(x)=2 x-3, g(x)=x3+5. The function ( text fog )-1(x) is equal to (A) ((x+7/2))1 / 3 (B) (x-(7/2))1 / 3 (C) ((x-2/7))1 / 3 (D) ((x-7/2))1 / 3

Tardigrade Admin · Accepted Answer

We have, f(x)=2 x-3, g(x)=x3+5 ( fog ) x=f(x3+5) =2(x3+5)-3=2 x3+7 Let y=( fog ) x=2 x3+7 ⇒ x=((y-7/2))1 / 3 ⇒ ( text fog )-1 x=((x-7/2))1 / 3

Q. Let $f : R \to R, g : R \to R$ be two functions such that $f (x) = 2 x - 3, g (x) = x^{3} + 5$ . The function $(fog)^{- 1} (x)$ is equal to

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

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

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

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

We have, $f (x) = 2 x - 3, g (x) = x^{3} + 5$
$($ fog $) x = f (x^{3} + 5)$
$= 2 (x^{3} + 5) - 3 = 2 x^{3} + 7$
Let $y = ($ fog $) x = 2 x^{3} + 7$
$\Rightarrow x = (\frac{y - 7}{2})^{1/3}$
$\Rightarrow (fog)^{- 1} x = (\frac{x - 7}{2})^{1/3}$

Q. Let f:R→R,g:R→R be two functions such that f(x)=2x−3,g(x)=x3+5. The function ( fog )−1(x) is equal to

(2x+7​)1/3

(x−27​)1/3

(7x−2​)1/3

(2x−7​)1/3