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The first derivative test is a method used in calculus to find the relative extrema (minima and maxima) of a function. This involves calculating the first derivative of the function to determine where it equals zero or where it's undefined. These values mark potential critical points.

When approaching a problem such as this one, calculating the first derivative is essential. Here, for function (a), the first derivative is calculated as follows: for \( f(x) = x^2 + \frac{k}{x} \), the derivative is \( f'(x) = 2x - \frac{k}{x^2} \). The task is to set \( f'(x) \) equal to zero to identify the critical points: \( 2x - \frac{k}{x^2} =0\).

Once the critical points are identified, the test helps determine whether each critical point is a relative maximum, minimum, or neither, by examining the sign changes in the derivative around these points.**

**Checking Function (b):** The derivative \( f'(x) = \frac{k - x^2}{(x^2 + k)^2} \) shows critical points when \( f'(x) = 0 \) or \( k - x^2 = 0 \). This helps find the values of \( x \) that are potential extremas.

Critical points are vital in determining the shape and behavior of the graph of a function. They occur when the first derivative is zero or undefined. In calculus optimization, these points are where a function's slope changes, indicating potential maxima or minima.

**Finding Critical Points for Function (a):** By setting the first derivative \( f'(x) = 2x - \frac{k}{x^2}\) to zero, the critical points are calculated as follows: \( 2x = \frac{k}{x^2} \). Taking \( x=3 \), we substitute into the equation to find \( k = 54 \). Hence, 3 is a critical point that needs further evaluation.

**Working with Function (b):** For \( f(x) = \frac{x}{x^2 + k} \), setting \( f'(x) = 0 \) offers another critical point value, \( k = x^2 \). Substituting \( x = 3 \) results in \( k = 9 \). Thus, these results from the derivative equations are the key milestones in assessing whether the critical point is a high or low peak on the curve.

The second derivative test provides further insight into the nature of the critical points found using the first derivative. It involves calculating the second derivative to see if it is positive, negative, or zero at each critical point identified. This test helps confirm whether a critical point is a relative minimum, relative maximum, or a point of inflection.

**Investigating Function (a):** The second derivative \( f''(x) = 2 + \frac{2k}{x^3} \) is evaluated at \( x = 3 \) with \( k = 54 \). Substituting these values gives \( f''(3) = 2 + \frac{108}{27} = 6 \), which is positive, indicating a relative minimum at the point where \( x=3 \).

**Function (b) Analysis:** For the second derivative of \( f(x) = \frac{x}{x^2 + k} \), the calculation shows \( f''(3) < 0 \) after simplifications when \( k = 9 \). This negative value signifies a relative maximum at the critical point \( x = 3 \).

The second derivative test, therefore, completes the examination of the function’s behavior around the critical points, confirming the nature of these critical points in the context of relative extrema.