Problem 31 - Calculus

Question

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

Step-by-Step Solution

Answer

(a) Increasing on \( (0, \infty) \), (b) never concave up.

Step 1: Calculate the Derivative

First, find the derivative of the function \( f(x) \). Since \( f(x) \) is a definite integral with a variable upper limit, we can use the Fundamental Theorem of Calculus. The derivative of \( f(x) = \int_{1}^{x} \frac{1}{\theta} \, d \theta \) with respect to \( x \) is \( f'(x) = \frac{1}{x} \).

Step 2: Determine Where the Function is Increasing

The function \( f(x) \) is increasing where its derivative is positive. Since \( f'(x) = \frac{1}{x} \), and \( x \) is always positive for \( x \geq 0 \), \( f(x) \) is increasing for \( x > 0 \).

Step 3: Calculate the Second Derivative

Determine concavity by calculating the second derivative of \( f(x) \). The second derivative of \( f(x) \), \( f''(x) \), is the derivative of \( f'(x) = \frac{1}{x} \). Using the power rule (differentiating \( \frac{1}{x} = x^{-1} \)), we get \( f''(x) = -\frac{1}{x^2} \).

Step 4: Determine Where the Function is Concave Up

A function is concave up where its second derivative is positive. Since \( f''(x) = -\frac{1}{x^2} \) and \( -\frac{1}{x^2} < 0 \) for all \( x > 0 \), \( f(x) \) is never concave up.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone in the world of calculus and provides a powerful link between differentiation and integration. It comprises two main parts:

• The first part states that if a function is continuous over an interval, then the function has an antiderivative, which can be expressed as a definite integral.
• The second part establishes that the derivative of an integral function is the original function itself, as long as the interval and function meet certain conditions.

In the exercise, where you have the function \( f(x) = \int_{1}^{x} \frac{1}{\theta} d \theta \), the Fundamental Theorem is applied to find the derivative of the integral. By using the theorem, it's determined that \( f'(x) = \frac{1}{x} \). Hence, the theorem greatly simplifies the process of finding derivatives of functions defined by integrals.

Derivative

Derivatives are essential in calculus for understanding how a function changes at any given point. They measure the rate of change or the slope of the function. In the context of the given exercise, calculating the derivative \( f'(x) \) from the integral is crucial to understanding where the function \( f(x) \) is increasing.
Finding the derivative of \( f(x) = \int_{1}^{x} \frac{1}{\theta} \, d \theta \) involves applying the Fundamental Theorem of Calculus. This results in \( f'(x) = \frac{1}{x} \). To determine where the function is increasing, you look for where this derivative is positive.
Since \( \frac{1}{x} \) is positive for all \( x > 0 \), the function \( f(x) \) is increasing on the interval \( x > 0 \). This shows how derivatives provide insight into the behavior of functions across different intervals.

Concavity

Concavity refers to the "bending" of the graph of a function and is determined by the second derivative. A function is concave up when the graph bends upwards, like a cup, and concave down when it bends downwards. To assess this, you observe the sign of the second derivative.
In the exercise, the second derivative \( f''(x) \) is derived from \( f'(x) = \frac{1}{x} \). Calculating this gives \( f''(x) = -\frac{1}{x^2} \). The sign of the second derivative determines the concavity of the graph.
Here, \(-\frac{1}{x^2}\) is negative for all \( x > 0 \), meaning that \( f''(x) \) is never positive. Hence, the function \( f(x) \) is always concave down for \( x > 0 \). Understanding concavity helps in visualizing the graph of a function and identifying points of inflection.