Math Problem Statement

how did you determine the local minima and local maxima

Solution

To determine local minima and local maxima, I follow a mathematical procedure rooted in calculus. Here's the step-by-step breakdown:


1. Find the First Derivative

  • The first derivative of a function f(x)f(x), denoted as f(x)f'(x), represents the slope of the function at each point.
  • To find critical points, I solve f(x)=0f'(x) = 0, because at these points, the slope of the tangent to the curve is zero (i.e., the function levels off).

2. Identify Critical Points

  • The solutions to f(x)=0f'(x) = 0 are potential candidates for local maxima, local minima, or inflection points.
  • I also consider points where f(x)f'(x) is undefined, as these could also indicate critical points.

3. Use the Second Derivative Test

  • To classify the critical points as local maxima or minima, I calculate the second derivative f(x)f''(x), which measures the concavity of the function:
    • If f(x)>0f''(x) > 0 at a critical point, the function is concave up, and the point is a local minimum.
    • If f(x)<0f''(x) < 0 at a critical point, the function is concave down, and the point is a local maximum.
    • If f(x)=0f''(x) = 0, the test is inconclusive, and I may use other methods like the First Derivative Test.

4. First Derivative Test (if necessary)

  • I examine the sign of f(x)f'(x) on intervals around the critical points:
    • If f(x)f'(x) changes from positive to negative at a critical point, it’s a local maximum.
    • If f(x)f'(x) changes from negative to positive at a critical point, it’s a local minimum.

5. Boundary Points (if applicable)

  • If the function is defined on a closed interval, I evaluate f(x)f(x) at the endpoints of the interval as well, since local extrema can occur there.

Example:

For the function f(x)=x33x2+4f(x) = x^3 - 3x^2 + 4:

  1. First derivative: f(x)=3x26xf'(x) = 3x^2 - 6x
  2. Solve f(x)=0f'(x) = 0: 3x(x2)=03x(x - 2) = 0, so x=0x = 0 and x=2x = 2.
  3. Second derivative: f(x)=6x6f''(x) = 6x - 6
    • At x=0x = 0: f(0)=6f''(0) = -6, so x=0x = 0 is a local maximum.
    • At x=2x = 2: f(2)=6f''(2) = 6, so x=2x = 2 is a local minimum.

Would you like further details, a worked-out example, or clarification on this method?

Follow-Up Questions:

  1. Can you show an example of finding local extrema for a specific function?
  2. What happens if f(x)=0f''(x) = 0 at a critical point?
  3. How do you find extrema for multivariable functions (e.g., f(x,y)f(x, y))?
  4. What is the significance of inflection points in this context?
  5. How does this process differ for non-polynomial functions, like trigonometric or exponential ones?

Tip:

Always check both the critical points and endpoints (if defined on a closed interval) when finding extrema!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Concavity
Optimization

Formulas

First Derivative: f'(x)
Second Derivative: f''(x)
Critical Points: f'(x) = 0

Theorems

Second Derivative Test
First Derivative Test

Suitable Grade Level

Grades 11-12 or College Level