Question

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

Explanation:

Step1: Find the derivative of the function

Let \( f(x)=x + \frac{k}{x}=x + kx^{-1}\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \( f^\prime(x)\) is:
\(f^\prime(x)=1 - kx^{-2}=1-\frac{k}{x^{2}}\)

Step2: Use the critical point condition

For a function to have a relative maximum (or minimum) at a point \(x = a\), the derivative at that point \(f^\prime(a)=0\) (critical point). We know that \(x=-2\) is a critical point, so we substitute \(x = - 2\) into \(f^\prime(x)\) and set it equal to 0:
\(f^\prime(-2)=1-\frac{k}{(-2)^{2}} = 0\)
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:
\(k = 4\)
We should also verify that this is a maximum. We can take the second - derivative \(f^{\prime\prime}(x)=\frac{2k}{x^{3}}\). Substitute \(x=-2\) and \(k = 4\) into the second - derivative: \(f^{\prime\prime}(-2)=\frac{2\times4}{(-2)^{3}}=\frac{8}{-8}=-1<0\), which means the function is concave down at \(x = - 2\), so \(x=-2\) is a relative maximum.

Answer:

\(k = 4\)