Problem 38 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=\left\\{\begin{array}{ll}{2 x+1,} & {x \leq-1} \\ {x^{2}-2,} & {x>-1}\end{array}\right. $$

Step-by-Step Solution

Answer

The critical points of the function are \(x=-1\) and \(x=0\). The function is increasing on the intervals \((-\infty, -1]\) and \((0, \infty)\) and is decreasing on the interval \((-1, 0)\). The function has a local maximum at \(x=-1\) and a local minimum at \(x=0\).

Step 1: Finding Critical Points

Finding the critical points for a function involves setting the derivative equal to zero, or undefined. Thus, you will need to first differentiate the function within its two defined intervals. For the interval \(x \leq -1\), \(f(x)=2x+1\), the derivative is \(f'(x) = 2\), which never equals zero and is never undefined. Therefore, there are no critical points for \(x \leq -1\). For the interval \(x > -1\), \(f(x)=x^2-2\), the derivative is \(f'(x) = 2x\). This equals zero when \(x = 0\) and is never undefined. Therefore the critical points are \(x=0\) and \(x=-1\), the latter due to the function definition change.

Step 2: Intervals of Increase or Decrease

To determine the intervals of increase or decrease, you will need to solve inequalities with the derivative. For the interval \(x \leq -1\), the derivative \(f'(x) = 2\) is positive, which points to an increasing function. For the interval \(x > -1\), you would analyze the derivative \(f'(x) = 2x\), which is negative for \(-1 < x < 0\), indicating a decreasing function, and positive for \(x > 0\), indicating an increasing function. Therefore, the function increases on the intervals \((-\infty, -1]\) and \((0, \infty)\), and decreases on the interval \((-1, 0)\).

Step 3: Applying First Derivative Test for Relative Extrema

Applying the First Derivative test, you look at the changes in sign of the derivative at the critical points. There are two critical points, \(x = -1\) and \(x = 0\). At \(x = -1\), the derivative changes from positive to negative, indicating a local maximum. At \(x = 0\), the derivative changes from negative to positive, indicating a local minimum. Therefore, the function has a local maximum at \(x=-1\) and a local minimum at \(x=0\).

Step 4: Confirming the Results with a Graph

You will need to graph the function \(f(x)\) on a graphing utility. The shape of the graph should confirm your previous calculations, with the function increasing on the intervals \((-\infty, -1]\) and \((0, \infty)\), decreasing on the interval \((-1, 0)\), a local maximum at \(x=-1\) and a local minimum at \(x=0\).

Key Concepts

Critical Points

Critical points occur where the derivative of a function is either zero or undefined. These points are crucial because they are candidates for potential highs and lows in the function's graph. In our exercise, to find critical points for the piecewise function f(x), we differentiated each segment. The equation \(f'(x) = 2\) in the interval \(x \leq -1\) has no zeros, hence no critical points. However, the derivative \(f'(x) = 2x\) for \(x > -1\) zeroes out at \(x = 0\). Additionally, although not a zero of the derivative, \(x = -1\) is considered a critical point owing to a change in the function's definition. Understanding where and why we obtain critical points is essential, as it sets the stage for analyzing the function's behavior around these points.

Intervals of Increase or Decrease

The intervals where a function increases or decreases are discovered by examining the sign of its derivative. If the derivative is positive, the function is rising; if negative, the function is falling.

For the specific function in our exercise:

• For \(x \leq -1\), \(f'(x) = 2\), which is always positive, suggesting the function is increasing on the interval \( (-\infty, -1] \).
• For \(x > -1\), we analyze \(f'(x) = 2x\). It's negative when \( -1 < x < 0 \) (decreasing) and positive for \( x > 0 \) (increasing).

By knowing how to solve these inequalities and examining the sign of the first derivative, one can identify the critical intervals that specify where the function's graph ascends or descends.

Relative Extrema

Relative extrema represent the peaks and valleys of a function's graph and are found at certain critical points. The First Derivative Test helps us identify these by looking at how the sign of the derivative changes at critical points.

In our function:

• A change from positive to negative derivative at \(x = -1\) signals a relative maximum.
• A change from negative to positive at \(x = 0\) indicates a relative minimum.

Distinguishing these relative extrema is important for understanding the overall shape and turning points of the function's graph.

Graphing Functions

Graphing functions allow us to visually confirm the behavior and extrema predicted by the analytical processes. After finding critical values, intervals of increase/decrease, and relative extrema, graphing illustrates how these parts come together. When graphing our piecewise function, we should see:

• An upward trend as \(x\) approaches \( -1 \) from the left.
• A peak at \(x = -1\) then a descent until \(x = 0\).
• Finally, an upward trend from \(x = 0\) onwards.

This step not only serves as a visual confirmation of our calculations but also helps us better understand and remember the function's behavior.