Question

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

Explanation:

Step1: Compute derivative of function

Let $f(x) = x + \frac{k}{x}$. The derivative is $f'(x) = 1 - \frac{k}{x^2}$.

Step2: Set derivative to zero at x=-2

At relative extremum, $f'(-2) = 0$. Substitute $x=-2$:
$1 - \frac{k}{(-2)^2} = 0 \implies 1 - \frac{k}{4} = 0$.

Step3: Solve for k

Rearrange: $\frac{k}{4} = 1 \implies k = 4$.

Step4: Verify maximum (second derivative test)

Second derivative: $f''(x) = \frac{2k}{x^3}$. Substitute $x=-2, k=4$:
$f''(-2) = \frac{2(4)}{(-2)^3} = \frac{8}{-8} = -1 < 0$, so it's a maximum.

Answer:

4