Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 27-32, find the interval(s) on which the graph of \(y=f(x), x \geq 0\), is (a) increasing, and (b) concave up. $$ f(x)=\int_{1}^{x} \frac{1}{\theta} d \theta, x>0 $$

Step-by-Step Solution

Answer

(a) The function is increasing on \((0, \infty)\); (b) the function is never concave up for \(x > 0\).

Step 1: Interpret the Function

You are given the function \( f(x) = \int_{1}^{x} \frac{1}{\theta} \, d\theta \). This represents the integral of \( \frac{1}{\theta} \) with respect to \( \theta \) from 1 to \( x \). The integral computes the natural logarithm function, so the function simplified is \( f(x) = \ln(x) \).

Step 2: Determine Where the Function is Increasing

To find where \( f(x) \) is increasing, calculate the derivative \( f'(x) \). Using the fundamental theorem of calculus, \( f'(x) = \frac{d}{dx}\left(\int_{1}^{x} \frac{1}{\theta} \, d\theta\right) = \frac{1}{x} \). An increasing function requires \( f'(x) > 0 \). Since \( \frac{1}{x} > 0 \) for all \( x > 0 \), \( f(x) \) is increasing on the interval \( (0, \infty) \).

Step 3: Determine Where the Function is Concave Up

To find the concavity, calculate the second derivative \( f''(x) \). Again differentiating, \( f''(x) = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \). For a function to be concave up, \( f''(x) > 0 \). Here, \( -\frac{1}{x^2} \) is always negative for \( x > 0 \), so \( f(x) \) is not concave up on \( (0, \infty) \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics, dealing with the calculation of integrals. Integrals are essentially the area under a curve, and they can be thought of as the reverse operation of derivatives. In the provided exercise, you are given a function defined by an integral: \[ f(x) = \int_{1}^{x} \frac{1}{\theta} \, d\theta \]This type of integral is known as a definite integral because it has set limits, from 1 to \( x \). The process of integration involves finding a function whose derivative is the given function inside the integral. Here, the integral computes the natural logarithm, giving us the function \( f(x) = \ln(x) \). By understanding and applying fundamental theorems of calculus, you can interpret and calculate functions defined by integrals, helping to solve real-world problems involving areas, volumes, and more. Integral calculus is a crucial component in the toolkit of any calculus student, bridging the gap between derivatives and the summation of infinitesimal parts.

Function Increase

In calculus, determining when a function is increasing or decreasing is essential for understanding its behavior. A function is said to be increasing if its derivative is greater than zero. The derivative of a function gives the slope of the tangent to the curve at any given point, indicating whether the function is moving upwards or downwards.For the function provided: \[ f(x) = \ln(x) \]The derivative is:\[ f'(x) = \frac{1}{x} \]For \( x > 0 \), \( \frac{1}{x} > 0 \) meaning that the function is increasing over the interval \( (0, \infty) \).

• A positive derivative indicates the function is rising.
• Since \( \frac{1}{x} \) remains positive, our function continues to rise without bounds for all positive \( x \).

Understanding the behavior of derivatives and their relation to the increase or decrease of a function can be powerful when predicting or explaining the behavior of mathematical and real-world phenomena.

Concavity

Concavity refers to the direction in which a function curves. A function can be either concave up or concave down, depending on the sign of its second derivative.When a function's second derivative is positive, the function is concave up, resembling a cup (or a smile). On the contrary, if the second derivative is negative, the function is concave down.Consider the function \( f(x) = \ln(x) \) from our exercise. Its second derivative is:\[ f''(x) = -\frac{1}{x^2} \]Since \( -\frac{1}{x^2} < 0 \) for all \( x > 0 \), the function is concave down on the interval \( (0, \infty) \).

• Negative second derivative suggests the function's slope is decreasing, thus turning or bending downwards.
• In our case, since the second derivative is always negative, \( f(x) \) is concave down for every positive \( x \).

Recognizing and analyzing concavity helps in understanding the nature of critical points and optimizing problems where maximum or minimum values are of interest.