4.5.6 Minimal and Maximum Value of Expressions

These problems can often be solved without calculus by recognizing key properties.

Key Insight

The minimum value of a square is 0!

Example

Problem: What is the minimum value of 2(x−1)2+32(x - 1)^2 + 3?

Solution: The minimum of (x−1)2(x - 1)^2 is 0, so the minimum is 2(0)+3=32(0) + 3 = 3

Tips

  • If the expression isn’t factored, complete the square first
  • For sin⁡\sin and cos⁡\cos: minimum = -1, maximum = 1
  • Only use derivatives if you can’t find a simpler form

Problem Set 4.5.6

Practice these problems. Type your answer and press Enter to check:

Min of f(x) = (x+2)² + 2
Min of y = 2x² + 3
Min of f(x) = 2x² + 4x + 2
Min of f(x) = x² − 2
Min of sin(2x) − 3
Min of y = x² + 4x at y =
Min of sin(3x) − 5
Min of y = x² + 2x − 3
Max of cos(3x) − 2
Max of 5 − cos(3x)
Min of y = 3x² + 4x
Range of y = −x⁴ + 4 is y ≤