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find the absolute maximum absolute minimum of the function F(x)=(x^2-12x+32)^(1/3) on the interval [3,10]

User Denski
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Answer:

  • minimum: -∛4 ≈ -1.58740
  • maximum: ∛12 ≈ 2.28943

Explanation:

The expression under the radical can be written in vertex form as ...

x^2 -12x +32 = (x -6)^2 -4

The function will be symmetrical about x=6. The minimum is at x=6 where we have ...

F(6) = ∛(-4)

The maximum is at the end of the interval farthest from the vertex, so is ...

F(10) = ∛((10 -6)² -4) = ∛12

The minimum is -∛4 ≈ -1.587; the maximum is ∛12 ≈ 2.289.

find the absolute maximum absolute minimum of the function F(x)=(x^2-12x+32)^(1/3) on-example-1
User Ken Gregory
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