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: Set the first derivative equal to zero at \(x = - 2\)

Since the function has a relative maximum at \(x=-2\), then \(y^\prime(-2)=0\).
Substitute \(x = - 2\) into \(y^\prime\): \(1-\frac{k}{(-2)^{2}}=0\).
Simplify the equation: \(1-\frac{k}{4}=0\).

Step3: Solve for \(k\)

Add \(\frac{k}{4}\) to both sides of the equation: \(1=\frac{k}{4}\).
Multiply both sides by \(4\) to get \(k = 4\).

Step4: 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:

\(k = 4\)