Question

for what value of k will x + \frac{k}{x} have a relative maximum at x = -2

Explanation:

Step1: Find the first derivative

Let \( y = x+\frac{k}{x}=x + kx^{-1}\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), the first derivative \(y^\prime=1-\frac{k}{x^{2}}\).

Step2: Use the critical - point condition

Since \(x = - 2\) is a critical point (where \(y^\prime = 0\)), substitute \(x=-2\) into \(y^\prime\).
We get \(0 = 1-\frac{k}{(-2)^{2}}\).

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Step3: Check the second - derivative (to confirm it's a maximum)

Find the second derivative \(y^{\prime\prime}=\frac{2k}{x^{3}}\).
Substitute \(x = - 2\) and \(k = 4\) into \(y^{\prime\prime}\), \(y^{\prime\prime}=\frac{2\times4}{(-2)^{3}}=\frac{8}{-8}=-1<0\).

Answer:

\(k = 4\)