Suppose that f : [a, b] → R is a continuous function that satisfies f (x) ∈ [a, b]for all x ∈ [a, b]. Prove that if f is differentiable on (a, b) and |f ′(x)| < 1 for all x ∈ (a, b),then f has exactly one fixed point. (Hint: We know that there is at least one fixed point byHomework 6. Try a proof by contradiction, and use the Mean Value Theorem.
Question
Suppose that f : [a, b] → R is a continuous function that satisfies f (x) ∈ [a, b]for all x ∈ [a, b]. Prove that if f is differentiable on (a, b) and |f ′(x)| < 1 for all x ∈ (a, b),then f has exactly one fixed point. (Hint: We know that there is at least one fixed point byHomework 6. Try a proof by contradiction, and use the Mean Value Theorem.
Solution
Let's prove that if a continuous function f : [a, b] → R satisfies f(x) ∈ [a, b] for all x ∈ [a, b], and if f is differentiable on (a, b) with |f'(x)| < 1 for all x ∈ (a, b), then f has exactly one fixed point.
To prove this, we will use a proof by contradiction and the Mean Value Theorem.
Assume, for the sake of contradiction, that f has more than one fixed point. Let's say f(c) = c and f(d) = d, where c and d are distinct fixed points in [a, b].
Since f is continuous on [a, b] and differentiable on (a, b), it follows from the Mean Value Theorem that there exists a point e ∈ (c, d) such that f'(e) = (f(d) - f(c))/(d - c).
Since f(c) = c and f(d) = d, we have f'(e) = (d - c)/(d - c) = 1.
However, this contradicts the given condition that |f'(x)| < 1 for all x ∈ (a, b). Therefore, our assumption that f has more than one fixed point must be false.
Hence, we can conclude that f has exactly one fixed point.
Note: The language used in the original text is English.
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