Question

Finding Parameter Values Show that \(f(x)=x^{2}+(a / x)\) cannot have a local maximum for any value of \(a .\)

Short answer

For any value of \(a\), the function \(f(x)=x^{2}+(a / x)\) does not have a local maximum, because the derivative \(f'(x)\) never equals zero for real \(x\).

Step-by-step solution

Differentiate the function

The first derivative of the function \(f(x)=x^{2}+(a / x)\) is obtained using the power rule for differentiation. The derivative of \(x^{2}\) is \(2x\), and the derivative of \(a / x\) is \(-a / x^{2}\). Thus, \(f'(x) = 2x - a / x^{2}\).

Find critical points

The critical points of a function occur where its derivative is equal to zero or undefined. So, set \(f'(x) = 0\). This gives \(2x - a / x^{2} = 0\). We can multiply through by \(x^{2}\) to get \(2x^{3} - a = 0\). Solving for \(x\), we get \(x^{3} = a / 2\) and thus \(x = (a / 2)^{(1/3)}\).

Conclusion The function cannot have a local maximum

Because the function \(f'(x) = 2x - a / x^{2}\) never equals zero for real \(x\) and any value of \(a\), \(f(x)\) cannot have a local maximum.

Key concepts

Differentiation

Differentiation is a foundational concept in calculus that deals with the analysis of how a function changes or varies. Essentially, it measures the sensitivity to change of the output of a function with respect to a change in its input. When differentiating a function, you are calculating its derivative, which provides the rate of change or slope of the function at any point.

When we have a function like the one provided, \( f(x)=x^{2}+(a / x) \), differentiation helps us find the rate at which \( f(x) \) changes with \( x \). We use the derivative, \( f'(x) \) to determine the nature of the graph of the function—whether it's increasing or decreasing and by how much at any given point. Understanding the derivative is crucial, as it lays the groundwork for analyzing functions and anticipating their behavior.

Critical Points in Calculus

In calculus, critical points are where the function's derivative is either zero or undefined, and they are important in determining where a function could have local maxima or minima. To identify these points, we look for values of \( x \) where \( f'(x)=0 \) or where \( f'(x) \) does not exist.

In the provided exercise, after finding the derivative \( f'(x) = 2x - a / x^{2} \), we set it to zero to look for potential critical points. We solve \( 2x - a / x^{2} = 0 \) for \( x \) to find where the function could possibly change from increasing to decreasing, which could indicate a local maximum. However, it turns out that \( f(x) \) in this case does not have a local maximum for any value of \( a \) because the derivative never equals zero given the nature of the equation derived. This shows that critical points are essential for understanding the extremities of a graph, but they do not always guarantee the existence of maxima or minima.

Power Rule for Differentiation

The power rule is a basic and efficient technique for differentiating functions that consist of a variable raised to a power. It states that if \( f(x) = x^{n} \), where \( n \) is any real number, then the derivative of \( f \) with respect to \( x \) is \( f'(x) = nx^{n-1} \).

For instance, in our exercise, we applied the power rule to differentiate the term \( x^{2} \) resulting in \( 2x \). Likewise, when differentiating \( a / x \) we rewrote it as \( ax^{-1} \) and applied the power rule to get \( -a / x^{2} \). By using the power rule, we are able to find the first derivative \( f'(x) \) quickly and accurately, which is crucial for analyzing functions and finding critical points. It simplifies the process of differentiation and is a tool that students frequently use in calculus when dealing with polynomial functions.