Problem 51 - Calculus: Early Transcendentals

Question

If \( f(x) = x^2 + 10 \sin x \), show that there is a number \( c \) such that \( f(c) = 1000 \).

Step-by-Step Solution

Answer

Using the Intermediate Value Theorem, there exists \( c \in (30,31) \) such that \( f(c) = 1000 \).

Step 1: Understand the Problem

We need to find a number \( c \) such that \( f(c) = 1000 \) for the given function \( f(x) = x^2 + 10 \sin x \). This means we are looking for a root of the equation \( x^2 + 10 \sin x = 1000 \).

Step 2: Rearrange the Equation

To find \( c \), we rearrange \( f(x) = x^2 + 10 \sin x = 1000 \) to \( x^2 + 10 \sin x - 1000 = 0 \). We need to show there exists at least one solution to this equation.

Step 3: Analyze the Behavior at Specific Values

Evaluate \( f(x) \) at some specific values. For example, calculate:\[ f(30) = 30^2 + 10 \sin(30) = 900 + 10 \cdot 0.5 = 905 \] and \[ f(31) = 31^2 + 10 \sin(31) = 961 + 10 \sin(31) \] approximately considering \( \sin(31) \approx \sin(30) = 0.5 \), \[ f(31) \approx 961 + 5. ]\] Hence, \( f(31) \approx 966 \).

Step 4: Assess the Function Continuity

The function \( f(x) = x^2 + 10 \sin x \) is continuous everywhere because it is the sum of continuous functions, \( x^2 \) and \( \sin x \).

Step 5: Apply the Intermediate Value Theorem

By the Intermediate Value Theorem, since \( f(x) \) is continuous on \([30,31]\) and \( f(30) < 1000 < f(31) \), there must exist some \( c \in (30,31) \) such that \( f(c) = 1000 \).

Key Concepts

Continuous Functions

Continuous functions are those that have no breaks, holes, or jumps in their graphs. In mathematical terms, a function \( f(x) \) is continuous at a point \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \). This means that as \( x \) approaches \( a \), the values of \( f(x) \) approach \( f(a) \) smoothly.
Continuous functions have several important properties:

• They can be drawn without lifting the pencil from the paper.
• If two functions \( f(x) \) and \( g(x) \) are continuous, their sum, difference, and product are also continuous.
• The quotient of two continuous functions is continuous wherever the denominator is not zero.

For our exercise, the function \( f(x) = x^2 + 10 \sin x \) is continuous because it is composed of the continuous functions \( x^2 \) and \( \sin x \). Adding or subtracting a constant doesn't affect continuity either.

Finding Roots

Finding roots involves determining the values of \( x \) that make a function equal to zero, also known as the solutions of the equation. In our context, it means finding \( c \) such that \( f(c) = 1000 \). To tackle such a problem, a common approach is to rearrange the equation into a root-finding format: \( x^2 + 10 \sin x - 1000 = 0 \). This allows us to think of the problem in terms of finding zeros of a new function: \( g(x) = x^2 + 10 \sin x - 1000 \).
Finding roots can sometimes be straightforward, especially if the function is simple, but it can also involve:

• Use of numerical methods like the bisection method, which is based on finding intervals \([a, b]\) where the function changes sign.
• Graphical methods, where intersections on the graph help approximate where roots might be.
• The Intermediate Value Theorem, which guarantees a root exists in an interval if the function takes opposite signs at the interval endpoints.

For \( f(x) = x^2 + 10 \sin x \), because it is continuous and \( f(30) < 1000 < f(31) \), the Intermediate Value Theorem confirms that a root must exist in the interval \([30,31]\).

Trigonometric Functions

Trigonometric functions like \( \sin x \), \( \cos x \), and \( \tan x \) are fundamental in mathematics. They relate angles of a triangle to the lengths of its sides and are periodic, meaning they repeat values in regular intervals. This periodic behavior makes them unique and widely applicable in various fields, such as physics, engineering, and signal processing.
Important properties of sine function \( \sin x \) are:

• It is periodic with a period of \( 2\pi \) (about 6.283), meaning \( \sin(x + 2\pi) = \sin x \).
• Its range is \([-1, 1]\), reflecting its maximum and minimum values.
• It is continuous, so it smoothly transitions between values.

In the given exercise \( f(x) = x^2 + 10 \sin x \), the term \( 10 \sin x \) adjusts the function’s values periodically. It slightly changes the otherwise predictable quadratic \( x^2 \), adding a sinusoidal variation. This makes \( f(x) \) interesting and useful in demonstrating the concept of continuity and the Intermediate Value Theorem in action.