Problem 6
Question
The graph of the function \(f\) defined by $$ f(t)=\left\\{\begin{array}{ll} t & \text { for } 0 \leq t<1 \\ 0 & \text { for } t=1 \end{array}\right. $$ is shown in Figure Ex. 12.1.6. It does not have a high point. Furthermore, it does not have a horizontal tangent; yet it satisfies all but one of the hypotheses of Rolle's Theorem. Which hypothesis of Rolle's Theorem does it not satisfy?
Step-by-Step Solution
Verified Answer
The function does not satisfy the continuity hypothesis on [0, 1].
1Step 1: Understand Rolle's Theorem
Rolle's Theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one number \( c \) in \((a, b)\) such that \( f'(c) = 0 \). We need to check these conditions for the function \( f(t) \).
2Step 2: Check Continuity on [0, 1]
The function \( f(t) = t \) for \(0 \leq t < 1\) and becomes \(0\) at \(t = 1\). We need to check if \( f \) is continuous on \([0, 1]\). Since at \( t = 1 \), \( f(t) \) abruptly changes from \( t \) to 0, there is a discontinuity at \( t = 1 \). Thus, the function is not continuous on the whole interval \([0, 1]\).
3Step 3: Verify Differentiability on (0, 1)
For \(0 < t < 1\), the function \( f(t) = t \) is differentiable with derivative \( f'(t) = 1 \). However, the differentiability at \( t = 1 \) is irrelevant because we are concerned with the open interval \((0, 1)\) where differentiability holds.
4Step 4: Examine Functional Values, f(0) and f(1)
Check if \( f(0) = f(1) \). We find that \( f(0) = 0 \) but \( f(1) = 0 \), thus they are equal. Nevertheless, due to discontinuity, Rolle's Theorem is not fully applicable.
5Step 5: Identify the Unmet Hypothesis
Rolle's Theorem requires continuity on the entire closed interval \([a, b]\). Since \( f(t) \) is not continuous on \([0, 1]\), the hypothesis of continuity on \([a, b]\) is not satisfied.
Key Concepts
ContinuityDifferentiabilityPiecewise FunctionsCalculus
Continuity
Continuity is a fundamental concept in calculus that concerns the smoothness of a function over an interval. For a function to be continuous on a closed interval \([a, b]\), it must not have any breaks, jumps, or holes. Simply put, you should be able to draw the graph of a continuous function on that interval without lifting your pencil.
- An important requirement for Rolle's Theorem is that the function must be continuous on the entire closed interval in question.
- In the original exercise, the function \( f(t) \) is not continuous at \( t = 1 \), as it makes an abrupt transition from \( t \) to 0, leading to a discontinuity.
Differentiability
Differentiability is the property of a function that allows it to have a derivative, meaning that it has well-defined tangent lines at each point within its domain. For a function to be differentiable, it must also be continuous, but not all continuous functions are differentiable.
- In the exercise, \( f(t) = t \) is differentiable on the open interval \(0, 1\).
- Despite being continuous on \(0 < t < 1\), differentiability isn't needed at the endpoints for Rolle’s Theorem; it just needs to exist in the open interval.
Piecewise Functions
Piecewise functions are defined by different expressions depending on the interval of the input value. They provide a way to describe functions that behave differently across different parts of their domains. These functions can be continuous or discontinuous, and they require careful attention to how each piece connects at their endpoints.
- The function \( f(t) \) in the exercise is a classic example of a piecewise function with different expressions on defined intervals.
- At \( t = 1 \), \( f(t) \) switches from the linear form \( t \) to a constant value \( 0 \), resulting in discontinuity.
Calculus
Calculus is a vast area of mathematics dealing with rates of change and accumulation. It is built upon concepts like limits, derivatives, and integrals. Rolle’s Theorem is a specific result in calculus that connects these ideas by stating conditions under which a function’s derivative must be zero at some point in an interval.
- Rolle’s Theorem requires continuity and differentiability along with \( f(a) = f(b) \) on the interval \(a, b\).
- In calculus, failure to meet these conditions means the theorem is not applicable, as seen in the exercise due to lack of continuity.
Other exercises in this chapter
Problem 5
Show that for the function $$ F(h)=(1+h)^{1 / h} \quad F^{\prime}(h)
View solution Problem 6
Find a cubic polynomial, \(p(x)=p_{0}+p_{1} x+p_{2} x^{2}+p_{3} x^{3}\) that matches \(f(x)=e^{-x}\) at \(a=0\)
View solution Problem 7
Assume that the sixth degree polynomials, \(S(x), C(x),\) and \(E(x)\) that match, respectively, \(\sin x, \cos x,\) and \(e^{x},\) are $$ \begin{array}{l} S(x)
View solution Problem 7
Draw the graph of a continuous function \(F\) with domain [1,5] that has a non- negative derivative at every point between 1 and \(5,\) and for which 5 is a loc
View solution