Math Problem Statement

If f is differentiable on [a, b] and f'(a) < λ < f'(b) for some λ ∈ ℝ, prove that there exists c ∈ (a, b) such that f'(c) = λ. Also, show that the function f(x) = x^x for x ≥ 0 and f(x) = (1 + x)^(-1/x) for -1 ≤ x < 0 cannot be the derivative of any function g defined on ℝ.

Solution

Let's break the problem into parts and provide clear explanations:

Part (a):

To Prove: If ff is differentiable on [a,b][a, b] and f′(a)<λ<f′(b)f'(a) < \lambda < f'(b) for some λ∈R\lambda \in \mathbb{R}, then there exists c∈(a,b)c \in (a, b) such that f′(c)=λf'(c) = \lambda.

Proof Outline:

  1. Define a new function: Consider g(x)=f(x)−λxg(x) = f(x) - \lambda x. Note that gg is differentiable because ff is differentiable.

  2. Derive g′(x)g'(x): g′(x)=f′(x)−λg'(x) = f'(x) - \lambda Hence, finding g′(x)=0g'(x) = 0 is equivalent to finding f′(x)=λf'(x) = \lambda.

  3. Apply the Extreme Value Theorem:

    • Since g(x)g(x) is continuous on [a,b][a, b], it attains a minimum at some c∈[a,b]c \in [a, b].
    • If c∈(a,b)c \in (a, b), then by Fermat's Theorem (necessary condition for extrema), g′(c)=0g'(c) = 0, i.e., f′(c)=λf'(c) = \lambda.

  4. Examine boundary points aa and bb:

    • At aa, g′(a)=f′(a)−λ<0g'(a) = f'(a) - \lambda < 0.
    • At bb, g′(b)=f′(b)−λ>0g'(b) = f'(b) - \lambda > 0.

    Thus, g′(x)g'(x) must cross zero at some c∈(a,b)c \in (a, b) by the Intermediate Value Theorem.

Conclusion: There exists c∈(a,b)c \in (a, b) such that f′(c)=λf'(c) = \lambda.

Part (b):

To Show: The function

\begin{cases} x^x, & x \geq 0, \\ (1 + x)^{-1/x}, & -1 \leq x < 0 \end{cases}$$ cannot be the derivative of any function $$g$$ defined on $$\mathbb{R}$$, i.e., $$g'(x) \neq f(x)$$ on $$[-1, \infty)$$. #### Key Steps: 1. **Behavior of $$f(x)$$ around $$x = 0$$:** - For $$x > 0$$, $$f(x) = x^x = e^{x \ln x}$$. As $$x \to 0^+$$, $$x \ln x \to 0$$, so $$f(x) \to 1$$. - For $$x < 0$$, $$f(x) = (1 + x)^{-1/x}$$. As $$x \to 0^-$$, we have $$(1 + x)^{-1/x} \to e$$ (using the exponential approximation). Hence, $$f(x)$$ is **discontinuous at $$x = 0$$**, with: $$\lim_{x \to 0^+} f(x) = 1 \quad \text{and} \quad \lim_{x \to 0^-} f(x) = e.$$ 2. **Implication:** - A derivative $$g'(x)$$ of any function $$g(x)$$ must be continuous (a basic property of derivatives). - Since $$f(x)$$ is discontinuous at $$x = 0$$, it cannot be the derivative of any function $$g(x)$$ on $$\mathbb{R}$$. **Conclusion:** The discontinuity of $$f(x)$$ at $$x = 0$$ implies that $$f(x)$$ is not a valid derivative. --- Let me know if you'd like detailed steps for each part or further clarification! --- ### Expanding Questions: 1. Why does $$g'(c) = 0$$ imply $$f'(c) = \lambda$$ in part (a)? 2. How does the Intermediate Value Theorem apply to the proof in part (a)? 3. What are the implications of the Extreme Value Theorem for $$g(x)$$? 4. Can you elaborate on the continuity of derivatives and its role in part (b)? 5. Why does $$\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$$ create a contradiction for $$g'(x)$$? --- ### Tip: Always check for continuity when dealing with derivatives since discontinuities immediately rule out differentiability!

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Math Problem Analysis

Mathematical Concepts

Differentiability
Intermediate Value Theorem
Extreme Value Theorem
Continuity
Discontinuity

Formulas

g(x) = f(x) - λx
g'(x) = f'(x) - λ
f(x) = x^x for x ≥ 0
f(x) = (1 + x)^(-1/x) for -1 ≤ x < 0

Theorems

Intermediate Value Theorem
Extreme Value Theorem
Fermat's Theorem
Continuity of Derivatives

Suitable Grade Level

Grades 11-12

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