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Math Help - Maxima, minima points of inflection

  1. #1
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    Maxima, minima points of inflection

    Hi not sure if I've got this right.

    Determine any points at which the gradient of y=x^3-3x^2-1 is zero, and detemine whether these are maxima, minima or points of inflection.

    3x^2 - 6x
    3x(x-2)

    x = 2

    (2)^3-3(2)^2-1

    (2, -5)

    (left of turning point) x = 1

    3(1)^2-6(1)= -3

    (1, -3)

    (right of turning point) x = 3

    3(3)^2-6(3)= -9

    (1, -9)

    I'm not sure if I've done this right, but if I were to answer this I believe (2, -5) is minima.
    Last edited by Dcoz; August 31st 2011 at 05:10 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Maxima, minima points of inflection

    What you've done is almost fine but you have to be more clear and careful with mathematical expressions like:
    Quote Originally Posted by Dcoz View Post
    3x^2 - 6x
    3x(x-2)

    x = 2
    When you read it like this, this doesn't make sense although you mean:
    3x^2-6x=0 \Leftrightarrow 3x(x-2)=0 \Leftrightarrow 3x=0 \ \mbox{or} \ x-2=0
    So we have two zero's x=0 and x=2

    To determine if you're dealing with a minimum or a maximum you can make a sign table (I guess you've done something like that):
    You get a maximum (0,-1) and a minimum (2,-5) (review this)

    To determine if there're points of inflection then you've to use the second derivative ...
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    Re: Maxima, minima points of inflection

    Thank you. I got (0, -1) for the maximum, but wasn't certain. Thank you also for correcting me on how I displayed the working out.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Maxima, minima points of inflection

    You're welcome! Have you found the points of inflection?
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    Re: Maxima, minima points of inflection

    I honestly thought that was the question 100% answered due to there being no points of inflection?
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    MHF Contributor Siron's Avatar
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    Re: Maxima, minima points of inflection

    There's one point of inflection:
    \frac{d^2y}{dx^2}=\frac{d(3x^2-6x)}{dx}=6x-6
    To determine if there're inflection points let \frac{d^2y}{dx^2}=0 so:
    6x-6=0 \Leftrightarrow x=1

    If we enter x=1 in the cubic function we get for the y-coordinate:
    y=-3

    So the point of inflection is:
    (1,-3)
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    Re: Maxima, minima points of inflection

    Thank You! You've been a great help.
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    MHF Contributor Siron's Avatar
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    Re: Maxima, minima points of inflection

    You're welcome!
    Also take a look at the graph of the function to compare with the answers you found (you can look at the graph @wolphram alpha or just enter the function in your calculator).
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    Re: Maxima, minima points of inflection

    I only have a scientific calculator. I'm currently studying engineering product design and didn't believe I would require a graphing calculator.
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  10. #10
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    Re: Maxima, minima points of inflection

    Ok, then you can take a look @ wolphram alpha (if you want), it's a very useful site.
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    Re: Maxima, minima points of inflection

    I've started using it quite a bit now. Just wish it could help with my physics questions :-(
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    MHF Contributor Siron's Avatar
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    Re: Maxima, minima points of inflection

    There's also the PHF for physics questions .
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    Re: Maxima, minima points of inflection

    Yeah, but unfortuately they don't seem to be as active as the friendly people over here and for that I thank you again!
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  14. #14
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    Re: Maxima, minima points of inflection

    You're welcome!
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  15. #15
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    Re: Maxima, minima points of inflection

    Be careful, just checking a point on either side of a critical point and seeing that they are lower than the value at the critical point does NOT prove that the critical point is a maximum! For example, if f(x)= -x^4+ \frac{15}{4}x^2+ 1, f'= -4x^3+ \frac{15}{2}x= -4x(x^2- \frac{5}{2}) so x= 0 is a critcal point. Now f(2)= f(-2)= -16+ 15= -1 while f(0)= 1. But (0, 1) is in fact, a local minimum, not maximum.

    We should use, instead, either the "first derivative test", checking that if x< 0, but close to 0, f'(x)= -4x(x^2- \frac{5}{2}) we have three negative factors so that f'(x) is negative, while if x> 0 but close to 0, f'(x)= -4x(x^2- \frac{3}{2} has two negative factors (-4and x^2- \frac{3}{2}) but one positive factor (x) and so f'(x) is positive. f is decreasing to left of 0 and then increasing- x= 0 is a minimum, or the "second derivative test", seeing that the second derivative is f''(x)=-12x^2+ \frac{15}{4} so that f''(0)= \frac{15}{4}> 0. That tells us that the first derivative is increasing at x= 0. Since the first derivative was 0 there, it must be negative to the left an positive to the right- i.e. (0, 1) is a minimum.

    For your example, f(x)= x^3- 3x^2- 1, f'(x)= 3x^2- 6x= 3x(x- 2) so that, as you say, x= 0 and x= 2 are critical points. We can see that if x< 2, but close to 2 (so x> 0) 3 and x are positive but x- 2 is negative. That means that f'(x) is negative to the left of x= 2 so the function is decreasing toward x. if x> 2 then x> 0 also so all three factors are positive. f' is positive to the right of x= 2 which means that the function is increasing away from x= 2. That tells us that x= 2 gives a minimum. Or we could use the "second derivative test": f''(x)= 6x- 6 so f''(2)= 12- 6= 6> 0. Again, that tells us that x- 2 gives a minimum.

    Of course, x= 0 is also a critical point. If x< 0 then x is also less than 2 so we have one positive factor (3) and two negative factors (x and (x- 2)) so f'(x) is positive for x< 0. For x> 0 but close to 0 (so x< 2) 3 and x are positive but x- 2 is still negative. That means that f'(x) is negative. We have f increasing toward x= 0 and decreasing awy from it- x= 0 gives a maximum. Or, f''(0)= 6(0)- 6= -6< 0 so, again, x= 0 gives a maximum.
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