Optimization Math Example 1

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Example 1

easy
Find the local maximum and minimum values of f(x)=x3−3x+2f(x) = x^3 - 3x + 2.

Solution

  1. 1
    Find f′(x)=3x2−3=3(x2−1)=3(x−1)(x+1)f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1).
  2. 2
    Set f′(x)=0f'(x) = 0: critical points at x=1x = 1 and x=−1x = -1.
  3. 3
    Find f′′(x)=6xf''(x) = 6x. At x=−1x = -1: f′′(−1)=−6<0f''(-1) = -6 < 0, so local maximum.
  4. 4
    At x=1x = 1: f′′(1)=6>0f''(1) = 6 > 0, so local minimum.
  5. 5
    Evaluate: f(−1)=−1+3+2=4f(-1) = -1 + 3 + 2 = 4 (local max); f(1)=1−3+2=0f(1) = 1 - 3 + 2 = 0 (local min).

Answer

Local maximum: f(−1)=4f(-1) = 4; local minimum: f(1)=0f(1) = 0
The second derivative test classifies critical points: negative second derivative means concave down (local max), positive means concave up (local min). Always evaluate the function at the critical point to get the actual max/min value.

About Optimization

The process of using derivatives to systematically find maximum or minimum values of a function over a domain.

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More Optimization Examples

Example 2 hard

A farmer has 200 m of fencing to enclose a rectangular field. What dimensions maximize the enclosed

Example 3 easy

Find the critical points of [formula] and classify them.

Example 4 medium

Find the absolute maximum and minimum of [formula] on [formula].

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Related Concepts

Derivative