Question

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

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

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

Answer 1

Answer: (B)

$-\frac{4\sqrt{2}}{7^3{3}^2}$

Differentiating the given equation, we get
$3y^2{y}^{\prime }-3y^{\prime }+1=0$
$\Rightarrow y^{\prime }(-10\sqrt{2})=-\frac{1}{21}$
Differentiation again we get $6y{y}^{\prime 2}+3y^2{y}^{\prime \prime }-3y^{\prime \prime }=0$
$\Rightarrow f^{\prime \prime }(-10\sqrt{2})=-\frac{6.2\sqrt{2}}{(21{)}^4}=-\frac{4\sqrt{2}}{7^3{3}^2}$