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Chapter 4: Problem 11

Which of the functions satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. $$f(x)=\sqrt{x(1-x)}, \quad[0,1]$$

Short Answer

Expert verified
The function satisfies the Mean Value Theorem on [0, 1].

Step by step solution

01

Understand the Mean Value Theorem Requirements

The Mean Value Theorem (MVT) states that if a function \( f \) is continuous on a closed interval \([a, b]\) and differentiable on an open interval \((a, b)\), then there exists at least one \(c\) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). So we must check \( f(x) \) for continuity and differentiability in the given interval \([0, 1]\).
02

Check Continuity over Interval

The function \( f(x) = \sqrt{x(1-x)} \) involves a square root, which is continuous everywhere it is well-defined and the expression under the square root is non-negative. Therefore, on the interval \([0, 1]\), \(f(x)\) is defined and continuous since \(0 \leq x \leq 1\) ensures non-negativity.
03

Check Differentiability on Open Interval

Differentiate \( f(x) = \sqrt{x(1-x)} \). First, let \( u = x(1-x) \). Then \( f(x) = \sqrt{u} \). Using the chain rule, \( f'(x) = \frac{1}{2\sqrt{u}}(1 - 2x) \).\( f'(x) \) is defined where \(u > 0\), i.e., \(x(1-x) > 0\), which includes the open interval \((0, 1)\). Thus, \( f(x) \) is differentiable on \((0, 1)\).
04

Conclusion about Satisfying MVT

Given that \( f(x) \) is continuous on \([0, 1]\) and differentiable on \((0, 1)\), it satisfies all the hypotheses of the Mean Value Theorem on the interval \([0, 1]\). Therefore, the function satisfies the MVT on the given interval.

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Key Concepts

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

Continuity of a Function
In calculus, continuity is an essential property of functions that is often checked before applying the Mean Value Theorem (MVT). A function is continuous on a closed interval \[a, b\] if there are no breaks, jumps, or gaps in its graph over this interval. For the function discussed here, \ f(x) = \sqrt{x(1-x)} \, continuity ensures that it is well-behaved over the specified interval of \ [0,1] \.
  • Continuity means that you can draw the graph of the function without lifting your pencil from the paper in the interval.
  • The function \ f(x) = \sqrt{x(1-x)} \ involves a square root, which is continuous where the expression inside it is non-negative.
  • On the interval \[0, 1\], the expression \ x(1-x) \ is non-negative because \ 0 \leq x \leq 1 \, so the function \ f(x) \ is continuous there.
Therefore, the continuity condition of the MVT is satisfied for this function in the interval \[0, 1\]. By ensuring continuity, we can guarantee that there are no breaks in our function that could invalidate the application of the MVT.
Differentiability of a Function
Differentiability is another crucial condition required by the Mean Value Theorem. If a function is differentiable over an interval, it means that it has a defined slope at every point within this interval. Differentiability requires the function to be smooth, having no sharp corners and no vertical tangent lines within the open interval \(a, b\).
  • In this exercise, we consider the function \ f(x) = \sqrt{x(1-x)} \ to ensure that it is differentiable on \(0, 1\).
  • We differentiate using the chain rule, treating the inner function \ u = x(1-x) \, to find \ f'(x) = \frac{1}{2\sqrt{x(1-x)}}(1 - 2x) \.
  • The derivative \ f'(x) \ exists and is well-defined when \ x(1-x) \ is greater than zero, which holds true for \(0, 1\).
Thus, we conclude that \ f(x) \ is differentiable on the open interval \(0,1\), allowing us to satisfy the differentiability requirement of the Mean Value Theorem.
Closed Interval in MVT
Understanding the concept of a closed interval is vital when working with the MVT. A closed interval \[a, b\] includes both of its endpoints, which is indicated by the square brackets. For the MVT to apply, a function must be continuous on this closed interval.
  • In our exercise, the interval \[0, 1\] includes both 0 and 1. This means we are considering every point from 0 to 1, inclusively.
  • Continuity is checked across this full interval, while differentiability is required within its interior, \(a, b\), meaning not including \ a \ or \ b \.
  • The inclusion of endpoints is crucial for verifying continuity, ensuring there are no sudden jumps at \ x = 0 \ or \ x = 1 \.
By ensuring \ f(x) = \sqrt{x(1-x)} \ is continuous on the closed interval \[0, 1\], and differentiable on the open interval \ (0, 1) \, we adhere to the conditions necessary for utilizing the Mean Value Theorem correctly.

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