Question

I need help please, been stuck on this question.

Answer 1

Answer:

D)     $\frac{1}{6}$

g¹(1) =  $\frac{1}{6}$

The inverse of the  function   $g(x) = \frac{x^{\frac{1}{3} }-1 }{2}$

Step-by-step explanation:

Step(i):-

Given that f(x) = (2x+1)³

 Let  y =  (2x+1)³

       $y^{\frac{1}{3} } =2x+1$

      $2x = y^{\frac{1}{3} } -1$

       $x = \frac{y^{\frac{1}{3} }-1 }{2}$

Step(ii):-

y = f(x) ⇒  x = f⁻¹ (y)

  ⇒ $f^{-1} (y) = \frac{y^{\frac{1}{3} }-1 }{2}$

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

The inverse of the given function  

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

Differentiating equation (i) with respective to 'x', we get

      $g^{l} (x) = \frac{1}{2} X \frac{1}{3} x^{\frac{1}{3} -1}$

     $g^{l} (x) = \frac{1}{6} x^{\frac{-2}{3} }$

Final answer:-

Put x=1

  $g^{l} (1) = \frac{1}{6} 1^{\frac{-2}{3} } = \frac{1}{6}$