Problem 44 - Essential Calculus Early Transcendentals

Question

(a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts ( d ) to sketch the graph of \(f .\) \(f(x)=e^{\arctan x}\)

Step-by-Step Solution

Answer

The graph has horizontal asymptotes at \(y = e^{\frac{\pi}{2}}\) and \(y = e^{-\frac{\pi}{2}}\). It is increasing everywhere, with inflection points at \(x = -1\) and \(x = 1\).

Step 1: Find Horizontal Asymptotes

Horizontal asymptotes occur when \(x\) approaches infinity or negative infinity. Evaluate \(\lim_{{x \to \infty}} f(x)\) and \(\lim_{{x \to -\infty}} f(x)\). The function is \(f(x) = e^{\arctan x}\), and as \(x \to \infty\), \(\arctan x \to \frac{\pi}{2}\). Thus, \(\lim_{{x \to \infty}} e^{\arctan x} = e^{\frac{\pi}{2}}\). Similarly, as \(x \to -\infty\), \(\arctan x \to -\frac{\pi}{2}\), so \(\lim_{{x \to -\infty}} e^{\arctan x} = e^{-\frac{\pi}{2}}\). Hence, the horizontal asymptotes are \(y = e^{\frac{\pi}{2}}\) and \(y = e^{-\frac{\pi}{2}}\).

Step 2: Find Vertical Asymptotes

Vertical asymptotes occur when \(x\) approaches a finite value causing \(f(x)\) to approach infinity. For \(f(x) = e^{\arctan x}\), there are no \(x\) values that cause \(\arctan x\) to be undefined. So, there are no vertical asymptotes.

Step 3: Determine Critical Points for Increasing/Decreasing Intervals

Take the derivative of \(f(x)\) to determine intervals of increase or decrease. Compute \(f'(x) = e^{\arctan x} \cdot \frac{1}{1 + x^2}\). Set \(f'(x) > 0\) for increasing intervals and \(f'(x) < 0\) for decreasing intervals. \(f'(x)\) is always positive since the exponential and the fraction are always positive. Therefore, \(f(x)\) is increasing on \((-\infty, \infty)\).

Step 4: Identify Local Maximum and Minimum Values

Since \(f(x) = e^{\arctan x}\) is increasing everywhere (from the derivative analysis), there are no local maxima or minima.

Step 5: Find Intervals of Concavity and Inflection Points

Compute the second derivative to determine concavity: \(f''(x) = e^{\arctan x} \cdot \frac{1}{1 + x^2} - e^{\arctan x} \cdot \frac{2x^2}{(1 + x^2)^2}\). Simplify to \(f''(x) = e^{\arctan x} \cdot \frac{1 - x^2}{(1 + x^2)^2}\). Set \(f''(x) = 0\): it simplifies to \(1 - x^2 = 0\), resulting in \(x = \pm 1\). The intervals of concavity are: concave up on \((-\infty, -1) \cup (1, \infty)\) and concave down on \((-1, 1)\). Inflection points are at \(x = -1\) and \(x = 1\).

Step 6: Sketch the Graph Using Information

Using the asymptotes, intervals of increase, and concavity, sketch the graph: draw horizontal asymptotes at \(y = e^{\frac{\pi}{2}}\) and \(y = e^{-\frac{\pi}{2}}\). Indicate that the graph is always increasing and changes concavity at \(x = -1\) and \(x = 1\). The graph is concave down between \(x = -1\) and \(x = 1\), and concave up elsewhere.

Key Concepts

Asymptotes

Asymptotes are lines that a graph approaches but never actually reaches. In the context of the function we are analyzing, which is \( f(x) = e^{\arctan x} \), we identified its horizontal asymptotes. These asymptotes occur as \( x \) approaches infinity or negative infinity.
To find horizontal asymptotes, we evaluate the limits of \( f(x) \) as \( x \) heads towards infinity both positively and negatively:

• As \( x \to \infty \), \( \arctan x \to \frac{\pi}{2} \). Thus, \( \lim_{{x \to \infty}} e^{\arctan x} = e^{\frac{\pi}{2}} \).
• As \( x \to -\infty \), \( \arctan x \to -\frac{\pi}{2} \). Therefore, \( \lim_{{x \to -\infty}} e^{\arctan x} = e^{-\frac{\pi}{2}} \).

This leads to horizontal asymptotes at \( y = e^{\frac{\pi}{2}} \) and \( y = e^{-\frac{\pi}{2}} \). Note that for vertical asymptotes, they typically occur where a function becomes undefined. However, since \( \arctan x \) is always defined for all \( x \), there are no vertical asymptotes for this function.

Increasing and Decreasing Intervals

To determine where a function is increasing or decreasing, we examine its first derivative. Let's consider the function \( f(x) = e^{\arctan x} \). We first find the derivative:\[ f'(x) = e^{\arctan x} \cdot \frac{1}{1 + x^2} \]This derivative helps us determine the nature of the intervals:

• The exponential function \( e^{\arctan x} \) is always positive.
• The fractional part \( \frac{1}{1 + x^2} \) is also always positive.

This means \( f'(x) \) is always positive for all values of \( x \), indicating that the function is increasing throughout its entire domain. Consequently, there are no decreasing intervals for this function.

Local Extrema

Local extrema refer to local maxima or minima within a function. These points occur where the derivative changes sign, signifying a change in the direction of the function. Since \( f'(x) \) is positive across its entire domain in our analysis of \( f(x) = e^{\arctan x} \), this means that the function is continuously increasing with no changes in direction.
As a result, there are no local maxima or minima on the graph of this function. It's important to note that while this function is always increasing, it doesn't imply reaching infinite or non-infinite values due to the horizontal asymptotes limiting its growth rate as \( x \) approaches positive or negative infinity.

Concavity and Inflection Points

Concavity describes whether a function curves upwards or downwards across its domain. We determine this by analyzing the second derivative. For our function \( f(x) = e^{\arctan x} \), the second derivative is:\[ f''(x) = e^{\arctan x} \cdot \frac{1 - x^2}{(1 + x^2)^2} \]To find inflection points, solve \( f''(x) = 0 \), which resolves to:

• \( 1 - x^2 = 0 \) \( \Rightarrow x = \pm 1 \)

These are the points where the concavity changes.
Now, let's determine the intervals of concavity:

• Concave up when \( f''(x) > 0 \): This happens when \( x^2 < 1 \), or specifically, when \( x > 1 \) or \( x < -1 \).
• Concave down when \( f''(x) < 0 \): This occurs when \(-1 < x < 1 \).

Hence, inflection points where the concavity transitions are at \( x = -1 \) and \( x = 1 \). As you approach these points, notice a change in the graph's curvature, which provides a complete picture of the function's behavior.