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burhan101

  • one year ago

Determine the extreme values of this function on the given interval. f(x)= 1+ (x+3)² ; -2<x<6

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  1. SithsAndGiggles
    • one year ago
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    Calc or pre-calc?

  2. burhan101
    • one year ago
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    pre-cal @SithsAndGiggles

  3. SithsAndGiggles
    • one year ago
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    Ok, that method's a bit less work. So you're given a parabola in vertex form: \[f(x)=1+(x+3)^2\] \(f(x)\) has vertex (an extreme value point) at \((-3,1)\). \(-3\) is not in the interval \(-2<x<6\). So, you have something that looks like this: |dw:1371696732887:dw| This means that only the endpoints of this interval will contain the extreme values.

  4. SithsAndGiggles
    • one year ago
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    Which means you have only to check \(f(-2)\) and \(f(6)\).

  5. burhan101
    • one year ago
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    where did you get (-3,1) from ? :S

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