Problem 62
Question
Continuity on Closed Intervals Let \(f\) be continuous and never zero on \([a, b] .\) Show that either \(f(x)>0\) for all \(x\) in \([a, b]\) or \(f(x)<0\) for all \(x\) in \([a, b] .\)
Step-by-Step Solution
Verified Answer
By contradiction, the assumption reveals that if the function takes both positive and negative values in \([a, b]\), it would violate the given that the function never equals zero in this interval. Hence, either \(f(x)>0\ for\ all\ x\) in \([a, b]\) or \(f(x)<0\ for\ all\ x\) in \([a, b]\).
1Step 1: Function values on endpoints
Firstly, observe the values of function \(f\) at the end points \(a\) and \(b\). Because every continuous function preserves the sign in a closed interval, \(f(a)\) and \(f(b)\) cannot have different signs.
2Step 2: Assume Invalidity of Statement
Assume the contrary, i.e. the function takes both positive and negative values in the interval. This would mean that there exists a \(c\) in \([a, b]\) such that \(f(c)=0\), which violates the given that the function is never zero in the interval.
3Step 3: Proof by contradiction
Our assumption leads to a contradiction. Therefore, it must be invalid, and thus either \(f(x)>0\ for\ all\ x\) in \([a, b]\) or \(f(x)<0\ for\ all\ x\) in \([a, b]\).
Key Concepts
Continuous FunctionsClosed Interval MethodProof by ContradictionFunction Sign Preservation
Continuous Functions
Understanding continuous functions is crucial when analyzing behavior over an interval. A continuous function is one where small changes in the input result in small changes in the output. In other words, if you were to draw the function without lifting your pencil from the paper, it would indicate continuity. Mathematically, for any point c within the domain of f, we have the limit as x approaches c of f(x) is equal to f(c).
This property is essential when we look at functions over a closed interval, such as [a, b]. A function that's continuous on a closed interval has no jumps, breaks, or holes within that range. It's this characteristic that allows for certain assertions about the function's values and behavior on the interval. For example, if a function is continuous and never zero on a closed interval, like in our exercise, the Intermediate Value Theorem guarantees that it cannot just switch signs without hitting zero.
This property is essential when we look at functions over a closed interval, such as [a, b]. A function that's continuous on a closed interval has no jumps, breaks, or holes within that range. It's this characteristic that allows for certain assertions about the function's values and behavior on the interval. For example, if a function is continuous and never zero on a closed interval, like in our exercise, the Intermediate Value Theorem guarantees that it cannot just switch signs without hitting zero.
Closed Interval Method
The Closed Interval Method involves examining the behavior of continuous functions within the bounds of a closed interval. A closed interval [a, b] includes its endpoints: it consists of all numbers from a to b, including a and b themselves. When you apply the Closed Interval Method for analyzing a function's sign, you start by checking the function's values at the endpoints of the interval.
If a continuous function is not zero at any point within a closed interval, as our textbook problem suggests, the function must maintain the same sign throughout the interval. This 'same sign' property is due to the function continuously transitioning from one point to another, ensuring no sign change unless the function crosses zero, which it does not in this case.
If a continuous function is not zero at any point within a closed interval, as our textbook problem suggests, the function must maintain the same sign throughout the interval. This 'same sign' property is due to the function continuously transitioning from one point to another, ensuring no sign change unless the function crosses zero, which it does not in this case.
Proof by Contradiction
The technique of Proof by Contradiction is a powerful tool in mathematics. It starts by assuming that the statement we want to prove is false. Then, we logically deduce the consequences of this assumption. If this leads to an inconsistency or contradiction, it implies that our original assumption must be wrong. Therefore, the statement we set out to prove is actually true.
In the textbook solution, we begin by assuming that the function f takes on both positive and negative values. This assumption leads to the conclusion that there must be a point where f equals zero. However, this contradicts our given information that f is never zero on the interval [a, b]. Therefore, our assumption is false, supporting the claim that f maintains a consistent sign over the whole interval.
In the textbook solution, we begin by assuming that the function f takes on both positive and negative values. This assumption leads to the conclusion that there must be a point where f equals zero. However, this contradicts our given information that f is never zero on the interval [a, b]. Therefore, our assumption is false, supporting the claim that f maintains a consistent sign over the whole interval.
Function Sign Preservation
The concept of Function Sign Preservation in continuous functions is a direct consequence of the absence of abrupt changes or breaks. If a function f is continuous on an interval and does not include zero, it means that f must preserve its sign throughout that interval.
This sign preservation is what our original exercise revolves around. The fact that function f is continuous and never zero on the closed interval [a, b] implies that f can't switch from positive to negative or vice versa, without crossing zero. Since crossing zero is not an option, the function must either remain positive throughout the interval or remain negative throughout it, thus preserving the sign.
This sign preservation is what our original exercise revolves around. The fact that function f is continuous and never zero on the closed interval [a, b] implies that f can't switch from positive to negative or vice versa, without crossing zero. Since crossing zero is not an option, the function must either remain positive throughout the interval or remain negative throughout it, thus preserving the sign.
Other exercises in this chapter
Problem 61
Multiple Choice \(\lim _{x \rightarrow 2^{-}} \frac{x}{x-2}\) \((A)-\infty \quad(B) \infty\) (C) 1 (D) \(-1 / 2 \quad(\mathrm{E})-1\)
View solution Problem 62
In Exercises \(59 - 62 ,\) find the limit graphically. Use the Sandwich Theorem to confirm your answer. $$\lim _ { x \rightarrow 0 } x ^ { 2 } \cos \frac { 1 }
View solution Problem 63
Free Fall A water balloon dropped from a window high above the ground falls \(y = 4.9 t ^ { 2 } \mathrm { m }\) in \(t\) sec. Find the balloon's (a) average spe
View solution Problem 63
Multiple Choice $$\lim _{x \rightarrow 0} \frac{\sin (3 x)}{x}=$$ (A) 1\(/ 3 \quad\) (B) 1 (C) 3 (D) \(\sin 3 \quad\) (E) does not exist
View solution