91Ó°ÊÓ

When tackling optimization problems, derivatives play a crucial role. In this exercise, our goal is to minimize the function \( f(x) = \frac{1}{x} + 4x^2 \). But how do we begin optimizing? The first step involves computing the derivative of the function, which tells us the rate at which \( f(x) \) is changing at any point \( x \). For the given function, the derivative is calculated as \( f'(x) = -\frac{1}{x^2} + 8x \). This derivative gives us insights into the slopes of the function's graph:

• If \( f'(x) > 0 \), the function is increasing at \( x \).
• If \( f'(x) < 0 \), the function is decreasing at \( x \).
• If \( f'(x) = 0 \), the function has either a maximum, a minimum, or possible inflection point at \( x \).

By analyzing \( f'(x) \), we are equipped to identify key points on the graph, known as critical points, where the nature of the function changes.

Critical points are the x-values where the derivative \( f'(x) \) equals zero, or where the derivative does not exist. These points are potential candidates for local maxima, minima, or inflection points. In optimization, we are particularly interested in points that minimize or maximize our function. From the derivative \( f'(x) = -\frac{1}{x^2} + 8x \), setting it to zero helps find these critical points: \[-\frac{1}{x^2} + 8x = 0\]By solving this equation, we find the critical point as \( x = \frac{1}{2} \).

• This critical point suggests that the slope of \( f(x) \) transitions, indicating potential minimization or maximization.
• Understanding these transition points allows us to apply further tests to definitively categorize each critical point.

Once we identify critical points, like \( x = \frac{1}{2} \), the next step is to determine their nature. Here, the second derivative test comes into play. The second derivative \( f''(x) \) lets us analyze the concavity of the function at a given critical point. For concavity:

• If \( f''(x) > 0 \), the function is concave up at \( x \), suggesting a local minimum.
• If \( f''(x) < 0 \), the function is concave down at \( x \), indicating a local maximum.

For the given function, the second derivative is calculated as:\[f''(x) = \frac{2}{x^3} + 8\]Testing at \( x = \frac{1}{2} \) yields:\[f''\left(\frac{1}{2}\right) = \frac{2}{(\frac{1}{2})^3} + 8 = 16 + 8 = 24\]Since \( 24 > 0 \), the function is concave up at \( x = \frac{1}{2} \), confirming a local minimum. The second derivative test successfully verifies the optimization of \( f(x) \), ensuring that \( x = \frac{1}{2} \) provides the smallest possible value of the original function.