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Chapter 2: Problem 56

Verify the intermediate value theorem (2.26) for \(f\) on the stated interval \([a, b]\) by showing that if \(f(a) \leq w \leq f(b),\) then \(f(c)=w\) for some \(c\) in \([a, b].\) $$ f(x)=-x^{3} $$

Short Answer

Expert verified
Yes, \(f(c) = w\) for \(c = 0\).

Step by step solution

01

Identify the Function and Interval

The function given is \(f(x) = -x^3\). We need to verify the Intermediate Value Theorem (IVT) for this function on the interval \([a,b]\). Here, we are considering the interval \([a, b]\).
02

Evaluate the Function at the Endpoints

Calculate \(f(a)\) and \(f(b)\) and ensure they satisfy \(f(a) \leq w \leq f(b)\), where \(w\) is a value between \(f(a)\) and \(f(b)\).Since no specific \([a, b]\) is given, assume typical values like \([-2, 2]\) for demonstration:- \(f(a) = f(-2) = -(-2)^3 = 8\)- \(f(b) = f(2) = -(2^3) = -8\)Now, choose \(w\) such that \(-8 \leq w \leq 8\). For example, take \(w = 0\).
03

Check Conditions of the Intermediate Value Theorem

The Intermediate Value Theorem states that if \(f\) is continuous on \([a, b]\), and \(f(a) \leq w \leq f(b)\), then there exists \(c \in [a, b]\) such that \(f(c) = w\).Given \(f(x) = -x^3\), which is a polynomial and hence continuous everywhere, the conditions of the IVT are satisfied.
04

Solve for c

Find \(c\) in \([-2, 2]\) such that \(f(c) = 0\).Solve \(-c^3 = 0\), which gives \(c^3 = 0\) or \(c = 0\).Since \(c = 0\) lies within the interval \([-2, 2]\), the condition \(f(c) = 0 = w\) is satisfied.

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Key Concepts

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

Continuous Function
A continuous function is a type of function that does not have any sudden jumps, breaks, or holes at any point within its domain. This means you can draw the graph of a continuous function without lifting your pen off the paper. Understanding continuous functions is crucial because many mathematical theorems, like the Intermediate Value Theorem (IVT), depend on this property.
  • In mathematics, continuous functions have outputs that change smoothly as the inputs change smoothly.
  • When evaluating if a function is continuous, check if it is smooth across its entire interval.
  • A function is continuous if it is continuous at every point in the domain.
In the context of the given exercise, the polynomial function \(f(x) = -x^3\) is continuous across all real numbers. This continuity is confirmed because polynomial functions are generally continuous over their entire domain.
Polynomial Function
A polynomial function is a type of mathematical expression that includes variables and coefficients, involving only non-negative integer exponents of the variables. These functions are structured in the form \(f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\), where \(a_n\) to \(a_0\) are constants, and \(n\) is a non-negative integer.
  • Polynomials are known for their simple and predictable behavior.
  • The degree of a polynomial tells us the highest power of the variable present in the polynomial.
  • Polynomial functions are smooth and continuous for all real numbers.
For the polynomial given in the exercise, \(f(x) = -x^3\), it fits the form of a simple cubic polynomial with a degree of three. Such functions, due to their continuous nature, make them ideal candidates for applying the Intermediate Value Theorem.
Endpoints Evaluation
Endpoints evaluation refers to computing the values of a function at the boundaries of a given interval. This is crucial when applying the Intermediate Value Theorem, as it requires comparing function values at these endpoints to any intermediate value being considered.
  • For the interval \([-2, 2]\), calculating \(f(a)\) and \(f(b)\) gives us the values at these endpoints.
  • Using the function \(f(x) = -x^3\), we find:
    • \(f(-2) = -(-2)^3 = 8\)
    • \(f(2) = -(2)^3 = -8\)
  • These calculations help identify the range of values \(w\) can take, ensuring that the Intermediate Value Theorem's conditions are satisfied.
By performing the endpoints evaluation, we can confirm the IVT applies as long as \(-8 \leq w \leq 8\). This ensures there is some point \(c\) in \([-2, 2]\) for which \(f(c) = w\).
Existence of Solutions
The Intermediate Value Theorem provides a guarantee for the existence of solutions within a specific interval. Given that a function is continuous on a closed interval \([a, b]\), and there exists a number \(w\) such that \(f(a) \leq w \leq f(b)\), the theorem asserts that there must be some \(c\) in \([a, b]\) for which \(f(c) = w\).
  • If a continuous function transitions from below \(w\) to above \(w\) within an interval, it must cross \(w\) at some point.
  • This crossing point is the solution we seek, often interpreted as the root or zero in numerical problems.
  • The IVT does not provide the exact location of \(c\), but it ensures one exists.
In the exercise, by setting \(w = 0\), the condition \(-8 \leq 0 \leq 8\) is met. Solving \(-c^3 = 0\) leads us to identify \(c = 0\). Hence, the existence of the solution \(c\) within the interval \([-2, 2]\) is confirmed, showcasing how the theorem operates in practice.

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Most popular questions from this chapter

Exercise \(27-32:\) Sketch the graph of tie piecewise-defined function \(f\) and, for the indicated value of \(a\), find each limit, if it exists: (a) \(\lim f(x)\) (b) \(\lim f(x)\) |c) \(\lim f(x)\) $$ f(x)=\left\\{\begin{array}{ll} 3 x & \text { if } x \leq 2 \\ x^{2} & \text { if } x>2 \end{array} \quad a=2\right. $$

Use the sandwich theorem to verify the limit. $$ \begin{aligned} &\lim _{x \rightarrow 0} x \sin (1 / x)=0\\\ &\text { (Hint: Use } f(x)=-|x| \text { and } g(x)=|x| \text { .) } \end{aligned} $$

Find all numbers at which \(f\) is continuous. $$ f(x)=\frac{\sqrt{9-x}}{\sqrt{x-6}} $$

Find all numbers at which \(f\) is continuous. $$ f(x)=\frac{x}{\sqrt[3]{x-4}} $$

Let \(n\) denote an arbitary integer. Sketch the graph of \(f\) and find \(\lim _{x \rightarrow n^{-}} f(x)\) and \(\lim _{x \rightarrow n^{+}} f(x)\). $$ f(x)=\left\\{\begin{array}{ll} x & \text { if } x=n \\ 0 & \text { if } x \neq n \end{array}\right. $$
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