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}}). [ \begin{align*} \frac{k}{4}&=1\ k&=4 \end{align*} ]

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)