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

Answer:

$k = 4$

Explanation:

Step1: Define the function

Let $y=x+\frac{k}{x}=x + kx^{-1}$.

Step2: Find the first - derivative

Using the power rule, $y^\prime=1 - kx^{-2}=1-\frac{k}{x^{2}}$.

Step3: Set the first - derivative equal to zero at the given point

Since the function has a relative maximum at $x = - 2$, we substitute $x=-2$ into $y^\prime$ and set $y^\prime = 0$. So, $1-\frac{k}{(-2)^{2}}=0$.

Step4: Solve for $k$

$1-\frac{k}{4}=0$. Add $\frac{k}{4}$ to both sides: $1=\frac{k}{4}$. Multiply both sides by 4 to get $k = 4$.