Problem 33 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=\frac{e^{x}-e^{-x}}{2} $$

Step-by-Step Solution

Answer

To sketch the graph of the function \(f(x) = \frac{e^x - e^{-x}}{2}\), we analyze the first and second derivatives. We find that there are no critical points or inflection points. The function is increasing and concave up on the interval \((-\infty, \infty)\). The graph has a horizontal asymptote at \(y = 0\) as \(x\) approaches both \(\infty\) and \(-\infty\). With this information, we can draw a smooth curve starting from the horizontal asymptote, passing through the origin, and remaining concave up while increasing toward the asymptote from above.

Step 1: Find the first derivative and critical points

First, take the derivative of the function \(f(x) = \frac{e^x - e^{-x}}{2}\) with respect to \(x\). Using the chain rule and linearity of derivatives, we have: \(f'(x) = \frac{e^x - (-e^{-x})}{2}\) Now, to find the critical points, let \(f'(x) = 0\): \(\frac{e^x - (-e^{-x})}{2} = 0\) \(e^x - (-e^{-x}) = 0\) \(e^x + e^{-x} = 0\) There are no real solutions for this equation, so there are no critical points.

Step 2: Find the second derivative and inflection points

Now, let's take the second derivative of the function \(f(x) = \frac{e^x - e^{-x}}{2}\). Again, using the chain rule and linearity of derivatives, we get: \(f''(x) = \frac{e^x + e^{-x}}{2}\) Now, to find the inflection points, let \(f''(x) = 0\): \(\frac{e^x + e^{-x}}{2} = 0\) \(e^x + e^{-x} = 0\) Similar to the first derivative, there are no real solutions for this equation, so there are no inflection points.

Step 3: Determine intervals of increasing/decreasing and concavity

From our analysis of the first derivative, we found that there are no critical points. Thus, we can determine the intervals of increasing/decreasing by noting that \(f'(x) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2}\) is always positive for all \(x\). Therefore, the function is increasing on the interval \((-\infty, \infty)\). From our analysis of the second derivative, we found that there are no inflection points. Thus, we can determine the intervals of concavity by noting that \(f''(x) = \frac{e^x + e^{-x}}{2}\) is also always positive for all \(x\). Therefore, the function is concave up on the interval \((-\infty, \infty)\).

Step 4: Identify asymptotes

Since our function is a rational function where the degrees of the numerator and denominator are equal, we can identify horizontal asymptotes by considering the leading terms of the numerator and denominator (which are both of degree 1). In this case: As \(x\) approaches \(-\infty\), the dominant term of the numerator is \(e^{-x}\), and the function approaches a horizontal asymptote of \(y = 0\). As \(x\) approaches \(\infty\), the dominant term of the numerator is \(e^x\), and the function approaches a horizontal asymptote of \(y = 0\). There is no vertical asymptote because the function is defined for all real \(x\).

Step 5: Sketch the graph

Based on the analysis above, we can sketch the graph of the function \(f(x) = \frac{e^x - e^{-x}}{2}\) using the following properties: - No critical points or inflection points - Increasing on the interval \((-\infty, \infty)\) - Concave up on the interval \((-\infty, \infty)\) - Horizontal asymptote at \(y = 0\) as \(x\) approaches both \(\infty\) and \(-\infty\) Following these guidelines, we can sketch a smooth curve starting from the horizontal asymptote and increasing while remaining concave up. The graph should pass through the origin and continue to increase as it approaches the horizontal asymptote from above.

Key Concepts

First Derivative Test

The first derivative test is a powerful tool in curve sketching that allows us to determine whether a function is increasing or decreasing at certain points. By taking the derivative of the function, we can analyze the critical points—where the first derivative is zero or undefined—to classify them as local maxima, minima, or neither.

For the function in our exercise, \( f(x) = \frac{e^x - e^{-x}}{2} \), we find the derivative to be \( f'(x) = \frac{e^x + e^{-x}}{2}\). When we set \(f'(x) = 0\), we realize that the expression \(e^x + e^{-x}\) cannot equate to zero, as the exponential function is always positive. Therefore, this function has no critical points, indicating it's always increasing or always decreasing. In this case, since the derivative here is always positive, the function is always increasing. This continuous increase means we don't have any relative maxima or minima.

Second Derivative Test

The second derivative of a function provides insight into its concavity, which refers to the direction the curve opens. If the second derivative is positive over an interval, the function is concave up, resembling a shape like \( \cup \). Conversely, if it is negative, it’s concave down (\( \cap \) shape).

To apply the second derivative test to our function \( f(x) = \frac{e^x - e^{-x}}{2} \), we compute the second derivative \( f''(x) = \frac{e^x + e^{-x}}{2} \). Since this expression is always positive, \( f''(x) > 0 \) for all \(x\), implying that our function is always concave up. This test doesn't provide us with any points of inflection here, since there are no points where the concavity changes. But it does confirm the overall

Increasing and Decreasing Functions

Understanding whether a function is increasing or decreasing over an interval can be tremendously helpful in graphing and interpreting the behavior of the function. An increasing function moves upwards as you move from left to right, while a decreasing function moves downwards.

In the case of our function \( f(x) = \frac{e^x - e^{-x}}{2} \), we determined that \( f'(x) > 0 \) for all \( x \), because the exponential function never takes on negative values. Consequently, \( f(x) \) is an always increasing function over \( (-\infty, \infty) \). This means that, as \( x \) increases or decreases, the value of \( f(x) \) consistently increases, and there are no intervals where it decreases.

Concavity and Inflection Points

The concavity of a curve tells us how the curve is curving, whether it's bending upwards or downwards. Inflection points are where the curve changes concavity—from concave up to concave down, or vice versa.

With our example function \( f(x) = \frac{e^x - e^{-x}}{2} \), the second derivative \( f''(x) = \frac{e^x + e^{-x}}{2} \), is positive for all \(x\), hence the function is concave up everywhere. This universal concavity means there are no inflection points, as the sign of \( f''(x) \) never changes. So, we would expect the graph of \( f(x) \) to always bend like a right-side up bowl, and this curvature will be present as the function extends to infinity in both the positive and negative directions.