Question

Let $f:(-1,1)\to R$ be the differentiable function with $f\left(0\right)=-1$ and $f^{\prime }\left(0\right)=1$.

If $g\left(x\right)={\left(f\right(2f\left(x\right)+2)}^2$, then $g^{\prime }\left(0\right)=$

• A. $0$
• B. $-2$
• C. $4$
• D. $-4$

Answer 1

Answer: (D)

$-4$

Given, $f\left(0\right)=-1,f^{\prime }\left(0\right)=1$

Also given $g\left(x\right)=f\left(2f\right(x)+2{)}^2$

Thus $g^{\prime }\left(x\right)=2f\left[\right(2f\left(x\right)+2\left)\right]2f^{\prime }\left(x\right)$

$\Rightarrow g^{\prime }\left(0\right)=2f\left[2f\right(0)+2]2f^{\prime }\left(0\right)$

$\Rightarrow g^{\prime }\left(0\right)=2[-1]2\left(1\right)=-4$