Problem 27 - Calculus with Concepts in Calculus

Question

Suppose \(f\) is differentiable at every number in \([-1,1]\), and \(f^{\prime}(-1)=1=f^{\prime}(1)\). Assume in addition that \(f(-1)=\) \(0=f(1)\). a. Explain why \(f\) has at least one zero in \((-1,1)\). b. Explain why \(f^{\prime}\) must have at least two zeros in \((-1,1)\).

Step-by-Step Solution

Answer

a. By the Intermediate Value Theorem, there must be a zero in \((-1,1)\). b. By Rolle's Theorem, there are at least two zeros of \(f'\) in \((-1,1)\).

Step 1: Analyze the Zero of Function

According to the given problem, the function \( f \) satisfies \( f(-1) = 0 \) and \( f(1) = 0 \). By the properties of continuous functions, if \( f \) is continuous on \( [-1, 1] \) and differentiable on \( (-1, 1) \) as given, then by the Intermediate Value Theorem, since it changes signs or remains the same at \( -1 \) and \( 1 \), there must be at least one \( c \) in \( (-1, 1) \) where \( f(c) = 0 \).

Step 2: Explore the Derivative Condition

Given that \( f'(-1) = 1 \) and \( f'(1) = 1 \), and \( f \) is differentiable across the interval \( [-1, 1] \), by Rolle's Theorem, there exists at least one \( c_1 \) in \( (-1, 0) \) where \( f'(c_1) = 0 \) and at least one \( c_2 \) in \( (0, 1) \) where \( f'(c_2) = 0 \). Rolle's Theorem applies here because \( f(-1) = f(1) = 0 \), so there are zeros of \( f' \) in each sub-interval.

Key Concepts

Rolle's Theorem

Rolle's Theorem is a useful concept in calculus that helps us identify whether a differentiable function has critical points or zeros of its derivative within a specific interval. It is especially powerful when dealing with differentiable functions bolstered by continuous behavior on closed intervals. Here's what we need to know about Rolle's Theorem:
The theorem states that if a function \( f \) is continuous over a closed interval \([a, b]\), differentiable over the open interval \((a, b)\), and \( f(a) = f(b) \), then there is at least one point \( c \) in \( (a, b) \) such that \( f'(c) = 0 \).

• This confers a real assurity that if a functional value repeats itself on both ends of a closed interval, then there's a change at least once in its slope somewhere in between.
• In simpler terms, if the function starts and ends at the same height, it must have been level or peaked/valleyed somewhere in between.

Applying this to the exercise, we know that \( f(-1) = 0 \) and \( f(1) = 0 \). So, by Rolle's Theorem, \( f'(c) = 0 \) must happen at least once in \((-1, 1)\).
This not only tells us that there is a point where the slope horizontally 'flattens' out but indicates that the function experiences a change indicative of a peak or trough within the interval.

Differentiable Functions

Differentiable functions are functions that have derivatives at all points within their domain, meaning they can be differentiated smoothly. A differentiable function will have a derivative that is a real number, which describes the rate at which the function's value changes at each point.

Why does this matter? Here are some essential characteristics:

• A differentiable function will not have any sharp corners or breaks, making it possible to draw a tangent line that represents the function’s slope at every point.
• Since it is smooth, the concept of a "derivative equals zero" means there's a point where the tangent line is horizontal, which are points often related to the function's maximums, minimums, or points of inflection where the function changes concavity.

To locate zeros of the derivative \( f'(c) = 0 \), it is critical that the function is smoothly changing and maintaining a form as would be expected in the exercise. We know \( f \) is differentiable over the interval \([ -1, 1 ]\), thus the slope is defined at all these points enabling us to apply the necessary calculus theorems like Rolle's.

Continuous Functions

A function is said to be continuous over a particular interval if there are no interruptions, jumps, or undefined points within it. Continuity is crucial because it ensures that every value in the interval has a point on the function.
If a function \( f \) is defined from \([-1, 1]\) and continuous on that interval, it implies:

• There are no sudden jumps or holes - the function can be drawn without lifting a pencil off the paper.
• The Intermediate Value Theorem can be applied, which states that if \( f(a) \) and \( f(b) \) have different values, every height between \( f(a) \) and \( f(b) \) must be reached at least once.

For the function \( f \) in the exercise, having \( f(-1) = 0 \) and \( f(1) = 0 \) with the functionality being continuous over \([-1, 1]\), ensures that all values, including zero, are achieved naturally, meaning somewhere within \((-1, 1) \), the function must cross the x-axis.
This guarantees that there is at least one point where \( f(c) = 0 \) in \((-1, 1)\) due to the alignment of continuity with the properties already given.