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Math Help - finding critical number for extrema

  1. #1
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    finding critical number for extrema

    im a little confused on this....here is my problem

    f(x)=(x+2)(x-1)

    f'(x) = 3x+6x critical numbers --- x=0 & x=-2 correct?
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  2. #2
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    Yes, x = -2 and x = 0 are the abs. of the two critical points.
    At x = -2, local max and at x = 0, local min.
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  3. #3
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    Quote Originally Posted by maybnxtseasn View Post
    im a little confused on this....here is my problem

    f(x)=(x+2)(x-1)

    f'(x) = 3x+6x critical numbers --- x=0 & x=-2 correct?
    Correct.
    Why are you confused ?
    When you differentiate it, you need to expand and collect like terms.
    And you did it correctly.
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  4. #4
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    dfs

    i thought i was but i got that one lol but this problem is getting me....

    f(x) = [x^(5)-5x]/5

    f'(x) = x-1

    i get x-1 was derivative which means critical point is x=1...back of book says critical point should be x= 1 & x= -1 ??? any help what i did wrong
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  5. #5
    Member Miss's Avatar
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    Quote Originally Posted by maybnxtseasn View Post
    i thought i was but i got that one lol but this problem is getting me....

    f(x) = [x^(5)-5x]/5

    f'(x) = x-1

    i get x-1 was derivative which means critical point is x=1...back of book says critical point should be x= 1 & x= -1 ??? any help what i did wrong
    You did not differentiate it correctly.
    You have f(x)=\frac{x^5-5x}{5}=\frac{1}{5}x^5-x
    then, f'(x)=x^4-1.
    Now, solve f'=0 \,.
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