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Calculating derivatives is a key part of solving optimization problems. To find the derivative of a function, you examine how the function changes as the input changes. In our exercise, we start with the function \[ y = x^2 + \frac{2}{x} \]Our task is to differentiate this function with respect to \( x \). We use the power rule for calculating the derivative of \( x^2 \) and the derivative of the term \( \frac{2}{x} \). The power rule states that if you have \( x^n \), then the derivative is \( nx^{n-1} \). For the term \( x^2 \), the derivative is \( 2x \). For \( \frac{2}{x} \), we convert it to \( 2x^{-1} \) and using the power rule, its derivative becomes \( -2x^{-2} \) or \( -\frac{2}{x^2} \). Thus, the derivative of the function is \[ y' = 2x - \frac{2}{x^2} \]This derivative tells us the rate at which \( y \) changes concerning changes in \( x \).

Critical points help us identify where a function's slope is zero or undefined, crucial for finding local maxima or minima. To find these points, we set the derivative equal to zero and solve for \( x \). In our case:\[ 2x - \frac{2}{x^2} = 0 \]By multiplying through by \( x^2 \), we eliminate the fraction:\[ 2x^3 - 2 = 0 \]Solving for \( x \), we get \( x^3 = 1 \), thus \( x = 1 \). This point, \( x = 1 \), is where the derivative is zero, which means the function's slope changes direction. To confirm whether this point is a maximum, minimum, or neither, we later examine its value in the function.

In calculus, absolute maxima are the highest points over a specific interval, whether critical points or endpoints. For our function, we check both. First, we evaluate the function at the critical point \( x = 1 \), and the endpoints of our interval, \( x = 1 \) and \( x = 4 \). - At \( x = 1 \): \[ y(1) = 1^2 + \frac{2}{1} = 3 \]- At \( x = 4 \): \[ y(4) = 4^2 + \frac{2}{4} = 16 + 0.5 = 16.5 \]After comparing these values, \( y(4) = 16.5 \) is the largest, meaning the absolute maximum over the interval \([1, 4]\) is \( 16.5 \) at \( x = 4 \). The critical point \( x = 1 \) gives a smaller value than the endpoint \( x = 4 \).

The power rule is an essential tool in differentiation. It's a simple and powerful rule used to find the derivative of any function of the form \( x^n \). According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \). This rule significantly simplifies the differentiation process, allowing us to quickly find derivatives without complex calculations.For example, in our function \( y = x^2 + \frac{2}{x} \), we apply the power rule to both terms individually:- For \( x^2 \), using the power rule, \( n = 2 \), so the derivative is \( 2x^{2-1} = 2x \).- For \( \frac{2}{x} = 2x^{-1} \), it becomes \( -2x^{-2} \) after differentiation.The power rule is especially useful in algebra-based calculus, where functions are often polynomials or monomials with integer exponents. Its straightforward nature allows us to handle complex problems with ease.