Question

If $\alpha$ is such a minimum value for which the inverse of $f(x)=x^2+3x-3$ exists in $[\alpha ,\infty )$ and $g$ is the inverse of the $f$ then ar $x=\alpha +\frac{5}{2},\frac{dg}{dx}=$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (D)

We have, $f(x)=x^2+3x-3$
$\Rightarrow y=x^2+3x-3$
$\Rightarrow y=x^2+3x-3+\frac{9}{4}-\frac{9}{4}$
$\Rightarrow y={\left(x+\frac{3}{2}\right)}^2-\frac{21}{4}$
$\Rightarrow y+\frac{21}{4}={\left(x+\frac{3}{2}\right)}^2$
$\Rightarrow x+\frac{3}{2}=\sqrt{y+\frac{21}{4}}$
$\Rightarrow x=\sqrt{y+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow f^{-1}(y)=\sqrt{y+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow f^{-1}(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}=g(x)$ (given)
Minimum value of $f^{-1}(x)$ is $\frac{-3}{2}$
$\therefore \alpha =-\frac{3}{2}$
$\therefore x=\alpha +\frac{5}{2}=-\frac{3}{2}+\frac{5}{2}=\frac{2}{2}=1$
Now, $g(x)=\sqrt{x+\frac{21}{4}}-\frac{3}{2}$
$\Rightarrow g^{\prime }(x)=\frac{1}{\sqrt{[2]}x+\frac{21}{4}}$
at $x=1,g^{\prime }(x)=\frac{1}{2\sqrt{1+\frac{21}{4}}}=\frac{1}{5}$