Question

In Exercises \(15-22,\) use analytic methods to find (a) the local extrema, (b) the intervals on which the function is increasing, and (c) the intervals on which the function is decreasing. $$h(x)=\frac{2}{x}$$

Short answer

The function \(h(x)=\frac{2}{x}\) has no local extrema. It is decreasing on the intervals \((-\infty,0)\) and \((0,\infty)\).

Step-by-step solution

Find the Derivative of the Function

The derivative of the function \(h(x)=\frac{2}{x}\) can be found using the power rule for derivatives, which states that the derivative of \(x^n\) is \(nx^{n-1}\). Here, \(n=-1\). So, the derivative \(h'(x)\) is \(-2x^{-2}\) or \(-\frac{2}{x^2}\).

Find Local Extrema

The local extrema of the function can be found by setting the derivative equal to zero and solving for \(x\). However, the equation \(-\frac{2}{x^2}=0\) has no solution, meaning the function \(h(x)=\frac{2}{x}\) has no local extrema.

Find Intervals Where the Function is Increasing or Decreasing

The function \(h(x)=\frac{2}{x}\) is increasing where its derivative is positive and decreasing where its derivative is negative. However, the derivative \(-\frac{2}{x^2}\) is always negative for \(x \neq 0\), meaning the function is always decreasing, except at \(x=0\) where it's not defined. Thus, the function is decreasing on the intervals \((-\infty,0)\) and \((0,\infty)\).

Key concepts

Local Extrema

To start, local extrema refer to the highest or lowest points on a graph within a small interval - essentially the peaks and troughs on a function's curve. In calculus, finding these points is pivotal because they give insights into the function's behavior and potential optimization problems.

Investigating the function given, which is
\( h(x) = \frac{2}{x} \), we determine the local extrema by first calculating its derivative and then setting this derivative equal to zero. However, in our case, the derivative
\( h'(x) = -\frac{2}{x^2} \) never equals zero. Thus, we conclude that no local extrema exist for \( h(x) \). Understanding this concept is crucial: just because a function exists, it doesn't guarantee that local extrema will be present.

Intervals of Increase and Decrease

Functions tell a story, and the part about intervals of increase and decrease relates to where the function is climbing up or sliding down. It's akin to plotting the course of a hill: you need to discern the ascents from the descents.

For the function
\( h(x) = \frac{2}{x} \), we look at the sign of the derivative
\( h'(x) = -\frac{2}{x^2} \). Since the derivative is negative for any real number not equal to zero, \( h(x) \) is always decreasing. Therefore, the entire real line, split by the undefined point at \( x = 0 \), forms two intervals: \( (-\infty, 0) \) and \( (0, \infty) \), both of which witness a steady decline in the function's value. Knowing these intervals helps predict how the function behaves without needing to graph it.

Derivative Applications

When it comes to calculus, derivatives aren't just a step in finding the slope; they're the powerhouse behind a vast array of applications in mathematics and beyond. They help in sketching curves, optimizing solutions in economics, and even tracking motion in physics.

In the context of our function
\( h(x) = \frac{2}{x} \), we applied the derivative
\( h'(x) = -\frac{2}{x^2} \) to determine the behavior of \( h(x) \). While we didn't find any local extrema, the derivative was instrumental in identifying the intervals of decrease. This instance shows how derivatives serve as a critical tool for analyzing functions. Students should appreciate their broad utility beyond simple slope calculations to effectively use calculus in solving real-world problems.