Math Problem Statement

Determine a continuous function f(x)f(x) and a real number aa such that:

f(x)f(x) has a critical point at x=ax=a.

f′(x)f′(x) is not defined at x=ax=a.

Both the left-hand limit and the right-hand limit of f′(x)f′(x) as x→axa exist and are finite.

The function f(x)f(x) has a point of inflection at x=ax=a.

Find such a function f(x)f(x) and specify the value of aa, providing a detailed explanation of your construction.

Solution

Problem Breakdown and Solution:

To satisfy the given conditions, we must construct a function f(x)f(x) such that:

  1. f(x)f(x)f(x)f(x) has a critical point at x=ax = a:

    • This implies f(x)f(x)+f(x)f(x)=2f(x)f(x)=0f'(x)f(x) + f(x)f'(x) = 2f(x)f'(x) = 0 at x=ax = a.
    • Thus, either f(x)=0f(x) = 0 or f(x)=0f'(x) = 0 at x=ax = a.
  2. f(x)f'(x) is not defined at x=ax = a:

    • This requires f(x)f(x) to have a point where f(x)f'(x) does not exist at x=ax = a.
  3. The left-hand and right-hand limits of f(x)f'(x) as xax \to a are finite:

    • This suggests a jump discontinuity in the derivative of f(x)f(x) at x=ax = a.
  4. f(x)f(x)f(x)f(x) has a point of inflection at x=ax = a:

    • This implies d2dx2[f(x)f(x)]=0\frac{d^2}{dx^2}[f(x)f(x)] = 0 changes sign at x=ax = a.

Function Construction:

We propose the piecewise function:

x, & x \leq 0, \\ x^2, & x > 0. \end{cases}$$ #### Verify Properties: 1. **Critical Point at $$x = a = 0$$:** - At $$x = 0$$, $$f(x) = 0$$, so $$f(x)f'(x) = 0$$. - This satisfies the critical point condition. 2. **$$f'(x)$$ not defined at $$x = 0$$:** - For $$x \leq 0$$, $$f'(x) = 1$$. - For $$x > 0$$, $$f'(x) = 2x$$. - Thus, $$f'(x)$$ is not defined at $$x = 0$$ due to a jump discontinuity. 3. **Left-hand and right-hand limits of $$f'(x)$$ as $$x \to 0$$:** - $$\lim_{x \to 0^-} f'(x) = 1$$. - $$\lim_{x \to 0^+} f'(x) = 0$$. - Both limits exist and are finite. 4. **Point of inflection at $$x = 0$$:** - For $$f(x)f(x) = f(x)^2$$: - $$\frac{d}{dx}[f(x)^2] = 2f(x)f'(x)$$. - For $$x < 0$$, $$f(x)^2 = x^2$$ (convex). - For $$x > 0$$, $$f(x)^2 = x^4$$ (concave for small $$x$$). - There is a change in concavity at $$x = 0$$, verifying the point of inflection. --- ### Final Answer: The function is: $$f(x) = \begin{cases} x, & x \leq 0, \\ x^2, & x > 0. \end{cases}$$ The value of $$a$$ is $$0$$. --- Would you like further clarifications? Here are some related questions to explore: 1. How can the concept of left and right derivatives be used in other contexts? 2. Can you derive an alternative piecewise function that satisfies similar conditions? 3. How does the second derivative test apply to points of inflection? 4. What is the geometric significance of jump discontinuities in derivatives? 5. How does this function relate to cusp-like behavior in calculus? **Tip:** When constructing piecewise functions, ensure continuity and evaluate all conditions for differentiability carefully.

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Math Problem Analysis

Mathematical Concepts

Critical points in calculus
Piecewise functions
Continuity and differentiability
Points of inflection
Jump discontinuities in derivatives

Formulas

Critical point condition: f'(x) = 0 or f'(x) is undefined
Inflection point condition: f''(x) changes sign

Theorems

Definition of critical points
Conditions for points of inflection
Piecewise function differentiability rules

Suitable Grade Level

Grades 11-12, College Level