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Math Help - Question about finding critical points in calculus?

  1. #1
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    Question about finding critical points in calculus?

    Okay, I'm stuck on this calculus problem. The section we're on covers monotonicity and the first derivative test, and we're supposed to be finding and classifying critical points in our homework problems. However, a couple of the problems I'm working on are different than the examples we did in class and I'm not sure exactly how to do them. Here's what I have so far:

    f(x) = 6x^3 - 9x^2 - 108x + 2
    f'(x) = 18x^2 - 18x - 108 <--- take the derivative
    0 = 18x^2 - 18x - 108 <--- set f(x) to 0 in order to solve for x

    Now, I tried using the quadratic formula and the website where we enter the answers said the ones I got were wrong. I could have just screwed something up... but does the quadratic formula even work for this type of problem?

    I kept going:
    108 = 18^2 - 18x
    108 = 18x ( x - 1 )

    The problem is all the examples we've done haven't had an integer left after we took the derivative, so rather than 108 we'd just have 0. If that were the case, the problem would be easy: the critical points would be 0 and 1. So what exactly do I need to do now? I don't understand what I'm supposed to be solving for (well, x, obviously, but I'm not sure how). Any suggestions would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by theant4 View Post
    Okay, I'm stuck on this calculus problem. The section we're on covers monotonicity and the first derivative test, and we're supposed to be finding and classifying critical points in our homework problems. However, a couple of the problems I'm working on are different than the examples we did in class and I'm not sure exactly how to do them. Here's what I have so far:

    f(x) = 6x^3 - 9x^2 - 108x + 2
    f'(x) = 18x^2 - 18x - 108 <--- take the derivative
    0 = 18x^2 - 18x - 108 <--- set f(x) to 0 in order to solve for x

    Now, I tried using the quadratic formula and the website where we enter the answers said the ones I got were wrong. I could have just screwed something up... but does the quadratic formula even work for this type of problem?

    I kept going:
    108 = 18^2 - 18x
    108 = 18x ( x - 1 )

    The problem is all the examples we've done haven't had an integer left after we took the derivative, so rather than 108 we'd just have 0. If that were the case, the problem would be easy: the critical points would be 0 and 1. So what exactly do I need to do now? I don't understand what I'm supposed to be solving for (well, x, obviously, but I'm not sure how). Any suggestions would be greatly appreciated!
    This factors nicely

    18x^2 - 18x - 108=0 \iff 18(x^2-x-6)=0 \iff 18(x-3)(x+2)=0
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  3. #3
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    Ah! Thank you very much! I've found that when I've been doing too much calculus I start to draw blanks when it comes to basic algebra :P
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