Problem 62 - Single Variable Calculus

Question

Use the Intermediate Value Theorem to find the value of \(c\) such that \(f(c)=M .\) \(f(x)=\frac{x-1}{x+1}\) on \([-4,-2] ; \quad M=2\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{x-1}{x+1}\) is continuous on the interval \([-4, -2]\) since the denominator is never zero within this interval. We have \(f(-4) = \frac{5}{3}\) and \(f(-2) = 3\). Since \(2 \in (\frac{5}{3},3)\), by the Intermediate Value Theorem, there exists a value \(c \in [-4, -2]\) such that \(f(c) = 2\). By solving the equation \(\frac{c-1}{c+1} = 2\), we find that \(c = -3\). Hence, \(f(-3) = 2\).

Step 1: Verify Continuity

First, we need to verify if the function f(x) = (x - 1) / (x + 1) is continuous on the interval [-4, -2]. A rational function is continuous everywhere except for where the denominator is zero. Since the denominator x + 1 is not equal to zero in the given interval (x = -1 is not in the interval [-4, -2]), the function is continuous on the interval.

Step 2: Calculate f(a) and f(b)

Now, we need to calculate the values of f(a) and f(b), where a = -4 and b = -2. 1. f(a) = f(-4) = \(\frac{(-4) - 1}{(-4) + 1}\) = \(\frac{-5}{-3}\) = \(\frac{5}{3}\) 2. f(b) = f(-2) = \(\frac{(-2) - 1}{(-2) + 1}\) = \(\frac{-3}{-1}\) = \(3\)

Step 3: Apply the Intermediate Value Theorem

We have f(a) = 5/3 and f(b) = 3. The given M=2, which is between f(a) and f(b) (5/3 < 2 < 3). Thus, the conditions for the Intermediate Value Theorem are satisfied and there must exist a value c in the interval [-4, -2] such that f(c) = 2.

Step 4: Solve for c

To find the value of c, we set f(c) to M, which is 2: \(f(c) = \frac{(c) - 1}{(c) + 1} = 2\) Now, solve for c: \(\frac{(c) - 1}{(c) + 1} = 2\) \((c-1) = 2(c+1)\) Expanding the equation: \(c - 1 = 2c + 2\) Moving all terms to one side: \(c - 2c = 2 + 1\) Simplifying: \(-c = 3\) Dividing by -1: \(c = -3\)

Step 5: Conclusion

There exists a value -3, which belongs to the interval [-4, -2], such that f(c) = 2. Therefore, c = -3 according to the Intermediate Value Theorem.

Key Concepts

Continuity of Functions

Continuity is a fundamental concept in calculus that indicates how a function behaves over an interval. A function is continuous if there are no jumps, breaks, or holes in its graph. For a formal definition: a function \( f(x) \) is continuous at a point \( x = c \) if the following conditions are met:

• \( f(c) \) is defined.
• The limit of \( f(x) \) as \( x \) approaches \( c \) exists.
• The limit equals the function's value at that point: \( \lim_{x \to c} f(x) = f(c) \).

When dealing with rational functions, which are ratios of two polynomials, the continuity is determined by the denominator. A rational function is continuous everywhere except at points where the denominator is zero, causing the function to be undefined.
In the problem above, the function \( f(x) = \frac{x-1}{x+1} \) is continuous on the interval \([-4, -2]\) because the denominator \(x+1\) is never zero within this interval.

Rational Functions

Rational functions are functions that can be expressed as the ratio of two polynomials. They have the general form:

• \( f(x) = \frac{P(x)}{Q(x)} \)

where \( P(x) \) and \( Q(x) \) are polynomials.
The key properties of rational functions include:

• They are continuous wherever the denominator is not zero.
• Vertical asymptotes occur at values that make the denominator zero, as long as these values do not cancel with factors in the numerator.
• They may also have horizontal or oblique asymptotes, depending on the degrees of the polynomials involved.

In the given exercise, \( f(x) = \frac{x-1}{x+1} \) is a rational function. It is continuous on the interval \([-4, -2]\) as \( x+1 eq 0 \) for any \( x \) in this set. Thus, continuity allows us to use the Intermediate Value Theorem to solve for \( c \) such that \( f(c) = M \).

Solving Equations

Solving equations involves finding the value(s) of the variable that make the equation true. In this exercise, we solve \( f(c) = 2 \) by manipulating the equation into a solvable form.

• Starting with \( f(c) = \frac{c-1}{c+1} = 2 \), we want to isolate \( c \).
• Multiply both sides by \( c+1 \) to remove the fraction: \((c-1) = 2(c+1)\).
• Expand and simplify the equation: \(c - 1 = 2c + 2\).
• Move terms involving \( c \) to one side: \(c - 2c = 2 + 1\).
• Simplify further: \(-c = 3\), so \(c = -3\) after dividing by \(-1\).

This step-by-step process reveals that \( c = -3 \) satisfies the equation, demonstrating how algebraic manipulation can solve rational equations effectively.