/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Determine whether the function i... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the function is continuous on the closed interval. \(h(t)=\frac{1}{t^{2}-9}, \quad[-2,2]\)

Short Answer

Expert verified
The function \(h(t) = \frac{1}{t^{2}-9}\) is continuous within the closed interval \([-2, 2]\) since the discontinuity points \(t = 3\) and \(t = -3\) are not within this interval.

Step by step solution

01

Identify the function and the interval

The function is given by \(h(t) = \frac{1}{t^{2}-9}\), and the closed interval to be considered is \([-2, 2]\).
02

Find the discontinuity points of the function

A rational function is discontinuous when the denominator becomes equal to zero. Let's find the values of t that make the denominator zero. \( t^{2} - 9 = 0 \) Now, solve the quadratic equation to find the values of t. \( t^{2} = 9 \) \( t = \pm \sqrt{9} \) \( t = \pm 3 \) So, the function \(h(t)\) will be discontinuous at \(t = 3\) and \(t = -3\).
03

Check if any discontinuity points are present within the closed interval [-2, 2]

Given that the discontinuity points are \(t = 3\) and \(t = -3\), neither of them is inside the closed interval \([-2, 2]\). It means that there is no discontinuity in the function within the given interval.
04

Conclusion

The function \(h(t) = \frac{1}{t^{2}-9}\) is continuous within the closed interval \([-2, 2]\) since none of its points of discontinuity are found within this interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Closed Interval
In mathematics, understanding the concept of closed intervals is crucial when analyzing function continuity. A closed interval, denoted by square brackets, includes both its endpoints. It is written as \[ a, b \] where the numbers a and b are included in the interval. For instance, the interval \[ -2, 2 \] includes all numbers from -2 to 2, including the endpoints themselves.

When discussing function continuity within such an interval, we verify if the function is continuous at every point in the interval, including the endpoints. This is important because continuity on an open interval (\( (a, b) \)) doesn't guarantee continuity at a and b. This subtle distinction can impact the applicability of certain theorems in calculus, like the Intermediate Value Theorem, which requires function continuity on a closed interval.
Rational Functions
A rational function is a fraction of two polynomials. It is in the form \( f(x) = \frac{p(x)}{q(x)} \) where \( p(x) \) and \( q(x) \) are polynomial functions, and \( q(x) \eq 0 \) since division by zero is undefined. One of the essential characteristics of rational functions relates to continuity. Typically, rational functions are continuous wherever the denominator is nonzero.

When analyzing rational functions like \( h(t) = \frac{1}{t^2-9} \), it is important to find where the denominator equals zero because these are the values for which the function will be undefined and may have discontinuities. Understanding where these discontinuities lie in relation to the interval in question is key to determining the continuity of the function over that interval.
Discontinuities
The concept of discontinuities in functions is pivotal when we wish to understand their behavior. Discontinuities are points where a function is not continuous. There are several types, including point, jump, and infinite discontinuities. Point discontinuity occurs if a function approaches different values from either side of a point, while jump discontinuity arises when there is a sudden 'jump' in function values from one side of a point to the other. Infinite discontinuity, which is particularly pertinent to rational functions, happens when function values increase or decrease without bound as they approach a particular point.

In our exercise, we look for values that make the denominator of a rational function equal zero, signaling potential infinite discontinuities. Since our function \( h(t) \) doesn't have discontinuities in the given closed interval \[ -2, 2 \], it remains continuous there.
Solving Quadratic Equations
Solving quadratic equations is a fundamental skill in algebra that also applies to analyzing function continuity. A quadratic equation takes the form \( ax^2 + bx + c = 0 \) and can be solved by various methods, including factoring, completing the square, the quadratic formula, or graphing. It is essential to identify the roots of the equation because they may represent x-values where the graph of a function intersects the x-axis or, in the context of rational functions, potential discontinuities.

In our example, the quadratic equation \( t^2 - 9 = 0 \) was factored as \( (t+3)(t-3) = 0 \) to find the roots t = 3 and t = -3. These values highlight where the function \( h(t) \) may not be continuous. However, since neither of these roots lies within the closed interval \[ -2, 2 \], the function remains continuous over that specific interval.

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