For what value of k will x+k/x have a relative max at x=-2
To find the value of \( k \) that will make the function \( f(x) = \frac{x + k}{x} \) have a relative maximum at \( x = -2 \), we can follow these steps:
Step 1: Compute the first derivative of \( f(x) \) with respect to \( x \).
The first derivative of \( f(x) \) can be found using the quotient rule. The quotient rule states that if we have a function \( g(x) = \frac{u(x)}{v(x)} \), the derivative \( g'(x) \) is given by:
\[ g'(x) = \frac{u'(x) \cdot v(x) - v'(x) \cdot u(x)}{v^2(x)} \]
Applying the quotient rule to \( f(x) = \frac{x + k}{x} \), we get:
\[ f'(x) = \frac{(1 \cdot x - 1 \cdot (x + k))}{x^2} \]
\[ f'(x) = \frac{x - (x + k)}{x^2} \]
\[ f'(x) = \frac{x - x - k}{x^2} \]
\[ f'(x) = \frac{-k}{x^2} \]
Step 2: Set the first derivative equal to zero and solve for \( x \).
To find the critical points of \( f(x) \), we set \( f'(x) = 0 \) and solve for \( x \).
\[ \frac{-k}{x^2} = 0 \]
Since the numerator, \(-k\), is equal to zero, the critical points \( x \) can be any real number except for \( x = 0 \).
Step 3: Determine the concavity of \( f(x) \) around the critical points.
To determine whether the critical point \( x = -2 \) is a relative maximum or minimum, we need to analyze the concavity of \( f(x) \) around this point.
To do this, we compute the second derivative of \( f(x) \) with respect to \( x \). The second derivative can be found by taking the derivative of \( f'(x) \):
\[ f''(x) = \frac{d}{dx}\left(\frac{-k}{x^2}\right) \]
\[ f''(x) = \frac{2k}{x^3} \]
Step 4: Substitute \( x = -2 \) into the second derivative expression.
Substituting \( x = -2 \) into the second derivative expression gives:
\[ f''(-2) = \frac{2k}{(-2)^3} \]
\[ f''(-2) = \frac{2k}{-8} \]
\[ f''(-2) = \frac{-k}{4} \]
Step 5: Analyze the sign of the second derivative.
The sign of the second derivative tells us about the concavity of the function. If the second derivative is positive, the function is concave up and has a local minimum at that point. If the second derivative is negative, the function is concave down and has a local maximum at that point.
Since we are looking for a relative maximum at \( x = -2 \), we need the second derivative to be negative at that point.
Step 6: Solve for \( k \) by setting \( f''(-2) \) less than zero.
Setting \( f''(-2) \) less than zero, we have:
\[ \frac{-k}{4} < 0 \]
To make the fraction negative, \( k \) must be positive. Therefore, any positive value of \( k \) will make the function \( f(x) = \frac{x + k}{x} \) have a relative maximum at \( x = -2 \).
In conclusion, for the function \( f(x) = \frac{x + k}{x} \) to have a relative maximum at \( x = -2 \), the value of \( k \) can be any positive number.
y = x + 4/x
has a max at x = -2