Answer: The approximate change in \(x\) is \(-\frac{1}{400}\), or \(-0.0025\).

Answer: The approximate change in \(x\) is \(-\frac{1}{400}\), or \(-0.0025\).

First find the value of \(x\) when \(y=4\):

\(\displaystyle 4=1+\frac2x+\frac1{x^2}\).

Multiplying by \(x^2\) gives

\(4x^2=x^2+2x+1\),

so

\(3x^2-2x-1=0\).

Factorising,

\((3x+1)(x-1)=0\).

Since \(x\gt 0\), \(x=1\).

Differentiate \(y=1+2x^{-1}+x^{-2}\):

\(\displaystyle \frac{dy}{dx}=-2x^{-2}-2x^{-3}\).

At \(x=1\),

\(\displaystyle \frac{dy}{dx}=-2-2=-4\).

For small changes,

\(\displaystyle \delta y\approx \frac{dy}{dx}\delta x\).

Here \(\delta y=0.01\), so

\(0.01\approx -4\delta x\).

Therefore

\(\displaystyle \delta x\approx -\frac{0.01}{4}=-\frac{1}{400}\).