Using intermediate value theorem show that a/(x-∆1) + b/(x-∆2) + c/(x-∆3) = 0 where a,b,c>0 & ∆1

To demonstrate that the equation \( \frac{a}{x - \Delta_1} + \frac{b}{x - \Delta_2} + \frac{c}{x - \Delta_3} = 0 \) has at least one solution using the Intermediate Value Theorem (IVT), we first need to understand the behavior of the function defined by the left-hand side of the equation. Let's denote this function as \( f(x) = \frac{a}{x - \Delta_1} + \frac{b}{x - \Delta_2} + \frac{c}{x - \Delta_3} \).

Understanding the Function's Behavior

Given that \( a, b, c > 0 \) and \( \Delta_1, \Delta_2, \Delta_3 \) are constants, we can analyze the function \( f(x) \) across its domain. The function is defined for all \( x \) except at the points \( x = \Delta_1 \), \( x = \Delta_2 \), and \( x = \Delta_3 \), where it has vertical asymptotes. This means that as \( x \) approaches any of these points from the left, \( f(x) \) will tend to \( -\infty \), and as \( x \) approaches from the right, \( f(x) \) will tend to \( +\infty \).

Behavior Near Asymptotes

Let’s examine the intervals around the asymptotes:

• As \( x \) approaches \( \Delta_1 \) from the left, \( f(x) \to -\infty \).
• As \( x \) approaches \( \Delta_1 \) from the right, \( f(x) \to +\infty \).
• As \( x \) approaches \( \Delta_2 \) from the left, \( f(x) \to -\infty \).
• As \( x \) approaches \( \Delta_2 \) from the right, \( f(x) \to +\infty \).
• As \( x \) approaches \( \Delta_3 \) from the left, \( f(x) \to -\infty \).
• As \( x \) approaches \( \Delta_3 \) from the right, \( f(x) \to +\infty \.

These observations indicate that in each interval between the asymptotes, the function \( f(x) \) transitions from negative to positive values. This is crucial for applying the Intermediate Value Theorem.

Applying the Intermediate Value Theorem

The IVT states that if a function is continuous on a closed interval \([a, b]\) and takes on different signs at the endpoints, then there exists at least one \( c \) in the interval \((a, b)\) such that \( f(c) = 0 \).

We can choose intervals based on the asymptotes:

• Consider the interval \( (\Delta_1 - \epsilon, \Delta_1 + \epsilon) \) for a small \( \epsilon > 0 \). Here, \( f(x) \) changes from \( -\infty \) to \( +\infty \).
• Similarly, for the intervals \( (\Delta_2 - \epsilon, \Delta_2 + \epsilon) \) and \( (\Delta_3 - \epsilon, \Delta_3 + \epsilon) \), the same behavior occurs.

In each of these intervals, since \( f(x) \) is continuous (except at the asymptotes) and takes on both negative and positive values, the IVT guarantees that there exists at least one \( x \) in each interval such that \( f(x) = 0 \).

Conclusion

Thus, we have shown that the equation \( \frac{a}{x - \Delta_1} + \frac{b}{x - \Delta_2} + \frac{c}{x - \Delta_3} = 0 \) has at least one solution in each interval between the asymptotes \( \Delta_1, \Delta_2, \) and \( \Delta_3 \). This is a powerful application of the Intermediate Value Theorem, illustrating how the behavior of rational functions can lead to the existence of roots in specified intervals.