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ksandoval

  • 3 years ago

how do i find the critical points of f(x) = (x-2)^5 (x+3)^4?

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  1. freewilly922
    • 3 years ago
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    How would you explain a critical point? A minimum/maximum? a zero?

  2. ksandoval
    • 3 years ago
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    hmm i dont know haha... i know that i need to find the derivative first though.

  3. freewilly922
    • 3 years ago
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    A critical point of a function of a single real variable, ƒ(x), is a value x0 in the domain of ƒ where either the function is not differentiable or its derivative is 0,

  4. freewilly922
    • 3 years ago
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    So it sounds like you need to take the derivative, set it to zero and then solve for x.

  5. ksandoval
    • 3 years ago
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    and i got \[f'(x) = 5(x-2)^4(x+3)^4 + 4(x+3)^3(x-2)^5\]

  6. ksandoval
    • 3 years ago
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    embarrassing, but i honestly have no idea how to factor that. ha.

  7. freewilly922
    • 3 years ago
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    Well, the idea is to find an alternative to brute force.

  8. freewilly922
    • 3 years ago
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    Ok, when things look ugly like this, substitute something easy. Like A= x-2 and B=x+3 This gives you \[f'(X) = 5A^4B^4 +4B^3A^5\] This is easy to factor and set to zero. Then once you have solved for when THIS is zero with respect to A and B then resubstitute the A=x-2 and B=x+3 to get you final answer.

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