Problem 61 - Calculus: Early Transcendentals

Question

Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. $$ y=\left(e^{x}+e^{-x}\right) / 2 $$

Step-by-Step Solution

Answer

The curve of \( y=\cosh(x) \) is symmetric about the y-axis, with a minimum point at \( (0, 1) \), no inflection points, and it grows to infinity in both directions without any asymptotes.

Step 1: Identify the Nature of the Function

The given function is an exponential function represented as \( y = \frac{e^x + e^{-x}}{2} \). This is known as the hyperbolic cosine function, \( y = \cosh(x) \), which is even and symmetric about the y-axis.

Step 2: Determine the Intercepts

For the y-intercept, set \( x = 0 \). We get \( y = \frac{e^0 + e^{0}}{2} = 1 \). Thus, the y-intercept is at \((0, 1)\). Since \( \cosh(x) \) is symmetric, it does not have x-intercepts as it never crosses the x-axis.

Step 3: Find the Derivative to Identify Critical Points

The first derivative of \( y = \cosh(x) \) is \( y' = \sinh(x) \). Critical points occur where \( y' = 0 \). However, \( \sinh(x) = 0 \) only at \( x = 0 \), which is already a known point (the y-intercept).

Step 4: Analyze the Second Derivative for Concavity and Inflection Points

The second derivative is \( y'' = \cosh(x) \). Since \( \cosh(x) > 0 \) for all \( x \), the curve is always concave up and does not have any inflection points.

Step 5: Check for Asymptotes

For horizontal asymptotes, examine the behavior as \( x \to \infty \) and \( x \to -\infty \). Both \( \lim_{x \to \infty} \frac{e^x + e^{-x}}{2} \) and \( \lim_{x \to -\infty} \frac{e^x + e^{-x}}{2} \) tend to infinity, hence there are no horizontal asymptotes. As a polynomial expression of exponentials, it has no vertical asymptotes either.

Step 6: Sketch the Curve

On a graph, plot the y-intercept \((0, 1)\). Draw a symmetric curve about the y-axis, always increasing and concave up (since \( \cosh(x) \) is positive and its second derivative is positive). The curve rapidly grows to positive infinity as \( x \) moves away from zero in both directions.

Key Concepts

Exponential Functions

Exponential functions are functions where the variable appears in the exponent. In the function \( y = \frac{e^x + e^{-x}}{2} \), the variable \( x \) is found in the exponents of \( e \) and its reciprocal.

This specific function is a type of hyperbolic function known as the hyperbolic cosine, denoted as \( \cosh(x) \). Hyperbolic functions are analogs of trigonometric functions but are related to the exponential function \( e^x \).

Key properties include:

• Symmetry: \( \cosh(x) \) is even and symmetric about the y-axis.
• Growth: The function increases rapidly, especially as \( x \) moves away from zero, due to the term \( e^x \).

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. For \( y = \cosh(x) \), the first derivative is \( y' = \sinh(x) \), which represents the hyperbolic sine function.

To find the critical points, solve \( \sinh(x) = 0 \). This equation only holds when \( x = 0 \).

In this case, the critical point \((0, 1)\) coincides with the y-intercept. Since \( \sinh(x) \) is the derivative of \( \cosh(x) \) and is only zero at \( x = 0 \), there are no other critical points to consider for this function.

Concavity

Concavity describes the "bending" of a curve. It can be determined by examining the second derivative of the function. For \( y = \cosh(x) \), the second derivative is \( y'' = \cosh(x) \).

Since \( \cosh(x) > 0 \) for all values of \( x \), the curve is always concave up.

Concave up curves are shaped like a cup, opening upwards, which indicates that the function grows faster as \( x \) increases. No "bending" change occurs, so there are no inflection points.

Intercepts

Intercepts are points where the graph intersects the axes. They are crucial for sketching the function.

Y-intercepts are found by setting \( x = 0 \). For \( y = \frac{e^x + e^{-x}}{2} \), this yields \( y = 1 \), leading to the y-intercept at point \((0, 1)\).

X-intercepts occur where \( y = 0 \), but since \( \cosh(x) \) is always positive, there are no x-intercepts.

These intercepts provide essential locations for graphing and understanding the behavior of the function. Remember, with this hyperbolic cosine function, the only intercept you'll find is on the y-axis.