Question

Consider the functions defined implicitly by the equation $y^3-3y+x=0$ on various intervals in the real line. If $xϵ(-\infty ,-2)\cup (2,\infty ),$ the equation implicitly defines a unique real valued differentiable function $y=f\left(x\right).$ If $xϵ(-2,-2)$ the equation implicitly defines a unique real valued differentiable function $y=g\left(x\right)$ satisfying $g=g\left(0\right)=0.$

If $f(-10\sqrt{2})=2\sqrt{2}$ then $f^{\prime \prime }(-10\sqrt{2})=$

• A. $\frac{4\sqrt{2}}{7^3\cdot 3^2}$
• B. $-\frac{4\sqrt{2}}{7^3\cdot 3^2}$
• C. $\frac{4\sqrt{2}}{7^3\cdot 3^3}$
• D. $\frac{4\sqrt{2}}{7^{}\cdot 3^{}}$

Answer 1

The correct option is B $-\frac{4\sqrt{2}}{7^3\cdot 3^2}$

$y^3-3y+x=0$
Or
$3y^2.y^{\prime }-3y^{\prime }+1=0$
Or
$3y^{\prime }(y^2-1)=-1$
Or
$y^{\prime }=\frac{1}{3(1-y^2)}$
Now
$y"=\frac{1}{3(1-y^2{)}^2}[-2y^{\prime }y]$
$=3\left(y^{\prime }{)}^2\right(-2y^{\prime }y)$
$=-6y{y}^{\prime 3}$
Now
$y^{\prime }=\frac{1}{3(1-y^2)}$ at $y=2\sqrt{2}$
$=\frac{1}{3(-7)}$
Hence
$y"=-12\sqrt{2}\frac{1}{3^3{.7}^3}$
$=\frac{-4\sqrt{2}}{3^2{.7}^3}$