Question

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

Explanation:

Step1: Define the function

Let \( f(x) = x + \frac{k}{x} \). We need to find the value of \( k \) such that \( f(x) \) has a relative maximum at \( x = -2 \).

Step2: Find the first derivative

To find critical points, we first find the derivative of \( f(x) \). Using the power rule, the derivative of \( x \) is \( 1 \), and the derivative of \( \frac{k}{x}=kx^{-1} \) is \( -kx^{-2}=-\frac{k}{x^2} \). So, \( f'(x) = 1 - \frac{k}{x^2} \).

Step3: Use the critical point condition

At a relative maximum (or minimum), the first derivative is zero (or undefined, but here \( x = -2 \) is in the domain where \( x
eq0 \), so we consider \( f'(-2) = 0 \)). Substitute \( x = -2 \) into \( f'(x) \):

\( f'(-2)=1 - \frac{k}{(-2)^2}=1 - \frac{k}{4} \)

Set \( f'(-2) = 0 \) (since at a relative extremum, the derivative is zero, provided the function is differentiable there, which it is for \( x
eq0 \)):

\( 1 - \frac{k}{4}=0 \)

Step4: Solve for \( k \)

Solve the equation \( 1 - \frac{k}{4}=0 \) for \( k \). Add \( \frac{k}{4} \) to both sides:

\( 1=\frac{k}{4} \)

Multiply both sides by \( 4 \):

\( k = 4 \)

Wait, but we also need to check if this is a maximum. Let's find the second derivative to confirm.

Step5: Find the second derivative

Differentiate \( f'(x)=1 - \frac{k}{x^2} \) with respect to \( x \). The derivative of \( 1 \) is \( 0 \), and the derivative of \( -\frac{k}{x^2}=-kx^{-2} \) is \( 2kx^{-3}=\frac{2k}{x^3} \). So, \( f''(x)=\frac{2k}{x^3} \).

Step6: Check the second derivative at \( x = -2 \)

Substitute \( x = -2 \) and \( k = 4 \) into \( f''(x) \):

\( f''(-2)=\frac{2(4)}{(-2)^3}=\frac{8}{-8}=-1 \)

Since \( f''(-2)= - 1<0 \), the function is concave down at \( x = -2 \), which means \( x = -2 \) is a relative maximum. So our value of \( k = 4 \) is correct. Wait, but let's re - examine the first step. Wait, maybe I made a mistake. Wait, the function is \( f(x)=x+\frac{k}{x} \), and we want a relative maximum at \( x=-2 \). Wait, let's go back.

Wait, when we found \( f'(-2)=0 \), we had \( 1-\frac{k}{4}=0\Rightarrow k = 4 \). But let's check the sign of the first derivative around \( x=-2 \). For \( x < - 2 \), say \( x=-3 \), \( f'(-3)=1-\frac{k}{9} \). If \( k = 4 \), \( f'(-3)=1-\frac{4}{9}=\frac{5}{9}>0 \). For \( x \) between \( -2 \) and \( 0 \), say \( x=-1 \), \( f'(-1)=1 - \frac{k}{1}=1 - k \). If \( k = 4 \), \( f'(-1)=1 - 4=-3<0 \). So the function is increasing before \( x=-2 \) and decreasing after \( x=-2 \) (for \( x \) in \( (- \infty,0) \)), so \( x=-2 \) is a relative maximum. So \( k = 4 \)? Wait, but let's check again. Wait, maybe I messed up the sign in the second derivative. Wait, \( f''(x)=\frac{2k}{x^3} \). If \( k = 4 \) and \( x=-2 \), then \( f''(-2)=\frac{2*4}{(-2)^3}=\frac{8}{-8}=-1<0 \), which confirms a maximum. So \( k = 4 \).

Wait, but let's start over. Let \( f(x)=x+\frac{k}{x} \), \( f'(x)=1-\frac{k}{x^2} \). Set \( f'(-2)=0 \):

\( 1-\frac{k}{(-2)^2}=0\Rightarrow1 - \frac{k}{4}=0\Rightarrow k = 4 \). Then, as we saw, the second derivative at \( x=-2 \) is negative, so it's a maximum. So the value of \( k \) is \( 4 \).

Answer:

\( \boxed{4} \)