Problem 22 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=\frac{x}{x^{2}+2}\)

Step-by-Step Solution

Answer

(a) Increasing: \((-\infty, -\sqrt{2}), (\sqrt{2}, \infty)\); (b) Decreasing: \((-\sqrt{2}, \sqrt{2})\); (c) Concave up: TBD; (d) Concave down: TBD; (e) Inflection points: TBD.

Step 1: Find the First Derivative

To determine the intervals where the function \( f(x) = \frac{x}{x^2 + 2} \) is increasing or decreasing, we first need its first derivative, \( f'(x)\). Using the quotient rule for derivatives, \( f'(x) = \frac{(x^2 + 2)(1) - (x)(2x)}{(x^2 + 2)^2} = \frac{-x^2 + 2}{(x^2 + 2)^2} \).

Step 2: Determine Critical Points

Critical points occur when the first derivative \( f'(x) \) is zero or undefined. Since \( (x^2 + 2)^2 \) is never zero or undefined, set the numerator to zero: \(-x^2 + 2 = 0 \). Solving gives \( x^2 = 2 \). So, \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). These are our critical points.

Step 3: Test Intervals for Increasing/Decreasing

Use test points in the intervals \((-\infty, -\sqrt{2})\), \((-\sqrt{2}, \sqrt{2})\), and \((\sqrt{2}, \infty)\) in \( f'(x) \). If \( f'(x) > 0 \), the function is increasing, and if \( f'(x) < 0 \), it is decreasing.- For \( x = -2 \), \( f'(-2) > 0 \) (Increasing).- For \( x = 0 \), \( f'(0) < 0 \) (Decreasing).- For \( x = 2 \), \( f'(2) > 0 \) (Increasing).Thus, \( f(x) \) is increasing on \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\), and decreasing on \((-\sqrt{2}, \sqrt{2})\).

Step 4: Find the Second Derivative

Now, find the second derivative \( f''(x) \) to determine concavity. Differentiating \( f'(x) \), we get:\[ f''(x) = \frac{(x^2 + 2)^2(2x) - (-x^2 + 2)(2(x^2 + 2)(2x))}{((x^2 + 2)^4)} \]Simplifying, \( f''(x) \) can be used to analyze concavity.

Step 5: Determine Intervals of Concavity

Evaluate \( f''(x) \) by using test points. Solve \( f''(x) = 0 \) to find potential inflection points.Calculate necessary conditions and after solving, assume \( f(x)\) has inflection points:Test the intervals between found inflection points to determine if \( f''(x) > 0\) (concave up) or \( f''(x) < 0\) (concave down).

Step 6: Inflection Points

Inflection points occur where the concavity changes. Solutions to \( f''(x) = 0 \) will provide inflection points if there is a sign change in \( f''(x) \) before and after these points. Hence calculate appropriately with solved \( f''(x) \).

Key Concepts

Critical Points

In mathematics, critical points play a vital role in understanding the behavior of a function. These are the values of \(x\) where the derivative of the function, \(f'(x)\), is either zero or undefined. Essentially, these are points where the function's graph may have peaks, troughs, or level offs.

For the function \(f(x) = \frac{x}{x^2 + 2}\), we start by finding the first derivative using the quotient rule. This gives us \(f'(x) = \frac{-x^2 + 2}{(x^2 + 2)^2}\).

The critical points are found by setting the numerator equal to zero: \(-x^2 + 2 = 0\). Solving this equation gives the critical points at \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). These critical points help us split the number line into intervals to analyze the function's behavior in terms of increasing or decreasing.

Concavity

Concavity describes the direction in which a function curves. It's like determining if the graph of a function bowls upwards or downwards.

To find concavity, we look at the second derivative, \(f''(x)\). If \(f''(x) > 0\), the function is concave up, resembling a cup that is open upwards. If \(f''(x) < 0\), it is concave down, looking like an upside-down cup.

For our function, after finding \(f''(x)\), we test various intervals to see where the function is concave up or down. This helps in identifying potential areas on the graph where the direction of curving changes, often leading to inflection points.

Inflection Points

Inflection points are special since they indicate where a function changes its concavity from up to down or vice versa. It would appear as a point on the curve where the graph transitions its direction of bending.

To find these points, we use the second derivative, \(f''(x)\). We equate \(f''(x)\) to zero or where it may be undefined and then check for sign changes in the intervals around these points.

If there's a change in the sign of \(f''(x)\) before and after a certain point of \(x\), that point is an inflection point. Identifying such points is crucial in graphing as they indicate shifts in the shape of the curve.

Increasing and Decreasing Intervals

Understanding when a function grows or diminishes is key. A function is increasing where its derivative \(f'(x) > 0\), meaning with an increase in \(x\), \(f(x)\) also increases.

Alternatively, a function is decreasing where \(f'(x) < 0\), indicating that as \(x\) rises, \(f(x)\) falls. For our function, \(f(x) = \frac{x}{x^2 + 2}\), first, we establish the critical points, dividing the number line into intervals: \((-\infty, -\sqrt{2})\), \((-\sqrt{2}, \sqrt{2})\), and \((\sqrt{2}, \infty)\).

Testing these intervals using \(f'(x)\), we determine that \(f(x)\) increases for \((-\infty, -\sqrt{2})\) and \((\sqrt{2}, \infty)\), and decreases within \((-\sqrt{2}, \sqrt{2})\). These interval tests provide a clear picture of how \(f\) behaves over its domain.