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Chapter 4: Problem 7

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ f(z)=\frac{1}{3}\left(z^{3}+z-4\right) ;[-1,2] $$

Short Answer

Expert verified
MVT applies; \(c = 1\) within \((-1, 2)\).

Step by step solution

01

Verify Continuity

The Mean Value Theorem (MVT) requires that the function be continuous on the closed interval \([-1, 2]\). The given function \(f(z) = \frac{1}{3}(z^3 + z - 4)\) is a polynomial function. Polynomial functions are continuous everywhere, so \(f(z)\) is continuous on \([-1, 2]\).
02

Verify Differentiability

The MVT requires the function to be differentiable on the open interval \((-1, 2)\). Polynomial functions are differentiable everywhere. Therefore, the function \(f(z)\) is differentiable on \((-1, 2)\).
03

Apply the Mean Value Theorem

The Mean Value Theorem states that if the function is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists at least one number \(c\) in \((a, b)\) such that:\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]For this function, \(a = -1\) and \(b = 2\).
04

Calculate \(f'(z)\)

First, we calculate the derivative of \(f(z)\):\[f'(z) = \frac{d}{dz}\left(\frac{1}{3} (z^3 + z - 4)\right) = \frac{1}{3}(3z^2 + 1) = z^2 + \frac{1}{3}.\]
05

Evaluate \(f(b)\) and \(f(a)\)

Evaluate \(f(2)\) and \(f(-1)\) to find the slope of the secant line:\[f(2) = \frac{1}{3}((2)^3 + 2 - 4) = \frac{1}{3}(8 + 2 - 4) = \frac{1}{3} \times 6 = 2.\]\[f(-1) = \frac{1}{3}((-1)^3 + (-1) - 4) = \frac{1}{3}(-1 - 1 - 4) = \frac{1}{3}(-6) = -2.\]
06

Solve for \(c\)

Substitute \(f(2)\) and \(f(-1)\) into the MVT equation:\[f'(c) = \frac{2 - (-2)}{2 - (-1)} = \frac{4}{3}.\]So, \(z^2 + \frac{1}{3} = \frac{4}{3}\).Solve this equation:\[ z^2 = \frac{4}{3} - \frac{1}{3} = 1\]\[ z = \pm 1\]However, \(-1\) is the boundary, only \(1\) is within \((-1, 2)\).
07

Conclude Possible Values of \(c\)

The Mean Value Theorem applies, and the only value of \(c\) in the interval \((-1, 2)\) that satisfies the theorem is \(c = 1\).

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Key Concepts

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

Polynomial Functions
A polynomial function is a mathematical expression involving a sum of powers of a variable multiplied by coefficients. These functions are expressed in the form:
  • Polynomials: \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\)
  • Example: \(f(z) = \frac{1}{3}(z^3 + z - 4)\)
Polynomial functions have some beautiful properties:
  • Continuous Everywhere: They do not have any breaks or holes.
  • Smooth Curves: Their graphs are smooth and without sharp corners.
This continuous nature plays a crucial role in some calculus theorems, such as the Mean Value Theorem. Knowing a function is polynomial helps us predict its behavior across an interval.
Continuity
Continuity is a property of a function that shows how smoothly it behaves over an interval. A function is continuous on an interval if, for any two points in the interval, the function can draw a line connecting these points that do not "jump". It can be expressed as:
  • Continuous Function: \(f(x)\) is continuous at \(x = a\) if: \(\lim_{x \to a} f(x) = f(a)\)
  • No Holes or Breaks: The graph is unbroken.
In this problem, the function is polynomial, which means it is continuous everywhere. This continuous behavior on the closed interval ([-1, 2]) ensures that the Mean Value Theorem can be applied because it requires continuity to hold.
Differentiability
Differentiability is a measure of how smooth a function's graph is, specifically how well-behaved the function is at each point in terms of having a tangent. A function is differentiable at a point if it has a defined derivative at that point. In other words:
  • Differentiability Condition: A function\( f \) is differentiable at \( a \) if \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) exists.
  • Smooth without Sharp Turns: The slope or gradient doesn't have abrupt changes.
For polynomial functions, differentiability is automatic because polynomials are smooth functions. In the open interval ((-1, 2)), our function \(f(z)\) satisfies differentiability, which meets another crucial requirement of the Mean Value Theorem.
Derivatives
Derivatives measure how a function changes as the input changes. It's like finding the slope of a function at any given point. The primary operation for determining a derivative is called differentiation, which provides us with insight into the function's critical points and rate of change. Regarding our problem:
  • Derivative of a Polynomial: For \(f(z) = \frac{1}{3}(z^3 + z - 4)\), the derivative is \(f'(z) = z^2 + \frac{1}{3}\).
  • Finding Critical Points: We use the derivative to identify where the slope of the function equals the slope of the line connecting two endpoints in our interval ([-1, 2]).
The Mean Value Theorem employs the derivative to assert that within the interval, the slope \(f'(c)\) matches the average slope \((f(2)-f(-1))/(2-(-1))=4/3\). Solving for \(c\) finds the exact point where this equality holds, which provides us a deeper understanding of how the function behaves within the interval.

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Most popular questions from this chapter

\text { Find } \int f^{\prime \prime}(x) d x \text { if } f(x)=x \sqrt{x^{3}+1}

What conclusions can you draw about \(f\) from the information that \(f^{\prime}(c)=f^{\prime \prime}(c)=0\) and \(f^{\prime \prime \prime}(c)>0\) ?

In each case, is it possible for a function \(F\) with two continuous derivatives to satisfy the following properties? If so sketch such a function. If not, justify your answer. (a) \(F^{\prime}(x)>0, F^{\prime \prime}(x)>0\), while \(F(x)<0\) for all \(x\). (b) \(F^{\prime \prime}(x)<0\), while \(F(x)>0\). (c) \(F^{\prime \prime}(x)<0\), while \(F^{\prime}(x)>0\).

Important news is said to diffuse through an adult population of fixed size \(L\) at a time rate proportional to the number of people who have not heard the news. Five days after a scandal in City Hall was reported, a poll showed that half the people had heard it. How long will it take for \(99 \%\) of the people to hear it?

Let \(E\) be a differentiable function satisfying \(E(u+v)=E(u) E(v)\) for all \(u\) and \(v\). Find a formula for \(E(x)\).
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