Problem 36
Question
A number \(a\) is called a fixed point of a function \(f\) if \(f(a)=a .\) Prove that if \(f^{\prime}(x) \neq 1\) for all real numbers \(x\) then \(f\) has at most one fixed point.
Step-by-Step Solution
Verified Answer
Since \( f'(x) \neq 1 \), \( f \) can have at most one fixed point.
1Step 1: Understanding the Problem
We need to prove that a function \( f \) has at most one fixed point, given that its derivative \( f'(x) eq 1 \) for all real numbers \( x \). A fixed point \( a \) satisfies \( f(a) = a \).
2Step 2: Assume Two Fixed Points
Assume for contradiction that \( f \) has two distinct fixed points, \( a \) and \( b \), such that \( f(a) = a \) and \( f(b) = b \).
3Step 3: Apply the Mean Value Theorem
By the Mean Value Theorem, since \( f \) is differentiable, there exists some \( c \) between \( a \) and \( b \) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \).
4Step 4: Calculate the Slope
Given \( f(a) = a \) and \( f(b) = b \), the expression becomes \( f'(c) = \frac{b - a}{b - a} = 1 \).
5Step 5: Contradiction Arises
Since \( f'(c) eq 1 \), there is a contradiction. Thus, \( f \) cannot have two distinct fixed points.
Key Concepts
Mean Value TheoremDifferentiable FunctionsDerivativeUnique Fixed Point
Mean Value Theorem
The Mean Value Theorem (MVT) is a fundamental concept in calculus which bridges the gap between derivatives and the behavior of functions over an interval. It specifies that for a function \( f \) that is continuous over a closed interval \([a, b]\) and differentiable over the open interval \((a, b)\), there exists at least one point \( c \) in \((a, b)\) where the slope of the tangent (derivative) at \( c \) is equal to the slope of the secant line joining the endpoints. In mathematical terms:
- \( f'(c) = \frac{f(b) - f(a)}{b - a} \)
Differentiable Functions
Differentiable functions are functions that have a derivative at every point in their domain. This means the function has a unique tangent at every point, and it implies the function is smooth without any sharp corners or edges.
- If a function is differentiable on an interval, it is also continuous on that interval.
- Not all continuous functions are differentiable (consider a function with a sharp corner).
Derivative
The derivative of a function represents the rate of change of the function's output with respect to changes in its input. In simpler terms, it's the slope of the function at any given point. The notation \( f'(x) \) is often used to denote the derivative of \( f \) at \( x \).
- A positive derivative indicates that the function is increasing.
- A negative derivative suggests the function is decreasing.
- If the derivative is zero, the function has a horizontal tangent (a potential maximum or minimum).
Unique Fixed Point
A unique fixed point is a value \( a \) for which \( f(a) = a \), and importantly, it is the only such value for the function within its domain. When a function satisfies the condition that its derivative is not equal to 1 everywhere (\( f'(x) eq 1 \)), it means any fixed point it has is unique.
- If two distinct fixed points existed, applying the Mean Value Theorem would imply the derivative equals 1, which contradicts our condition.
- Thus, ensuring \( f'(x) eq 1 \) secures the function against any more than one fixed point.
Other exercises in this chapter
Problem 36
(a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d
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\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=12+4 x-x^{2}, \quad[0,5]$$
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The illumination of an object by a light source is directly proportional to the strength of the source and inversely proportional to the square of the distance
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