Problem 74 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the relative extrema, if any, of each function. Use the second derivative test, if applicable. $$ g(x)=x^{2}+\frac{2}{x} $$

Step-by-Step Solution

Answer

The function \(g(x) = x^2 + \frac{2}{x}\) has a relative minimum at \(x = 1\).

Step 1: Find the first derivative of function

To find the critical points of the function, we first need to find the first derivative. Differentiate the function \(g(x) = x^2 + \frac{2}{x}\) with respect to x. Using the power rule and knowing that the derivative of \(\frac{2}{x}\) is \(-\frac{2}{x^2}\), \(g'(x) = 2x - \frac{2}{x^2}\)

Step 2: Find the critical points

To find the critical points where relative extrema might occur, we need to find the values of x for which the first derivative is equal to 0 or undefined. In this case, set \(g'(x)\) equal to 0 and solve for x. \(2x - \frac{2}{x^2} = 0\) Now, solve for x: 1. Multiply through by \(x^2\): \(2x^3 - 2 = 0\) 2. Divide through by 2: \(x^3 - 1 = 0\) 3. Factoring: \((x-1)(x^2 + x + 1) = 0\) The real solution to the equation is \(x = 1\), so this is a critical point.

Step 3: Find the second derivative of the function

In order to apply the second derivative test, we need to find the second derivative of the function. Differentiating \(g'(x)\) once more with respect to x, we get: \(g''(x) = 2 + \frac{4}{x^3}\)

Step 4: Apply the second derivative test

To determine if the critical point yields a relative extrema, evaluate the second derivative at the critical point: \(g''(1) = 2 + \frac{4}{1^3} = 2 + 4 = 6\) Since the second derivative is positive at \(x = 1\), the function is concave up at this point, indicating that the critical point is a relative minimum.

Step 5: State the answer

By analyzing the first and second derivatives, we've determined that there is a relative minimum at the point \(x = 1\). Thus, the relative extrema for the function \(g(x) = x^2 + \frac{2}{x}\) is a minimum at \(x = 1\).

Key Concepts

Relative Extrema

In the context of calculus, relative extrema refer to the local maximum or minimum points on a graph. These points are where the function reaches a peak or a trough respectively in a certain neighborhood or interval.
To identify these points, we typically use the first and second derivatives of the function. The first derivative helps find critical points, and the second derivative can confirm whether these points are maxima, minima, or saddle points.
For example, in the function \(g(x) = x^2 + \frac{2}{x}\), the relative extrema is found at the critical point \(x = 1\) by using the second derivative test, revealing a relative minimum.

Critical Points

Critical points are crucial in understanding where a function may have relative extrema, sharp turns, or points of inflection. These points occur where the first derivative of a function is zero or undefined.
For \(g(x) = x^2 + \frac{2}{x}\), the first derivative \(g'(x)\) was calculated to be \(2x - \frac{2}{x^2}\). Setting this equation to zero yields potential critical points. In this case, solving \(2x - \frac{2}{x^2} = 0\) results in the critical point \(x = 1\).
Identifying these points allows for further analysis with the second derivative to determine the nature of the extrema.

Concavity

Concavity describes how a function curves and helps differentiate between relative maxima and minima. For a function \(f(x)\), it's described as concave up if the graph opens upwards, similar to a cup, and concave down if it opens downwards.
The second derivative \(g''(x) = 2 + \frac{4}{x^3}\) was used to assess concavity at critical points. Evaluating \(g''(1) = 6\), which is positive, shows that the function is concave up at \(x = 1\). Thus, the critical point is a relative minimum.
Visualizing concavity gives insight into the function's behavior around critical points, confirming the type of extrema.

Power Rule

The power rule is a fundamental derivative rule that makes differentiation straightforward when dealing with expressions in the form of powers of \(x\). The rule states: if \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\).
In our example \(g(x) = x^2 + \frac{2}{x}\), the power rule was used to determine the derivative. The term \(x^2\) differentiated to \(2x\), using the power rule effectively. Meanwhile, \(\frac{2}{x}\) was rewritten as \(2x^{-1}\) and differentiated to \(-2x^{-2}\).
Applying the power rule simplifies finding derivatives and is an essential technique every student should grasp.