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Derivatives, inflections, concavity [included pictures]

  1. Nov 26, 2008 #1
    [​IMG]

    7. First I took derivative and got f'(x)= 6x^2-72x+210. divided the whole thing by 6 and found the factors for x^2-12x+35. (x-7)(x-5) which will give me that the smaller critical number is x=5 and the bigger critical number is x=7.

    8. First I set the equation to 0 and used the quadratic formula to get that x= -0.94721 and x= -0.05279. Afterwards, I plugged it back into the original equation and got that x=22.90486 which will be absolute maximum and x= 9.541833 which will be the absolute minimum. is this correct?

    9. f(x)= x^2 * e^8x , first i took f'(x) and got f'(x)= e^8x * 2x + x^2 * 8e^8x

    next , i took the second derivative and got f"(x)= [2x*8e^8x+2e^8x]+[8e^8x*2x + x^2*64e^8x]...

    the next step, i have to set it to zero? Can anyone help me with that and tell me if i am doing this correctly please. Thanks a lot.
     
  2. jcsd
  3. Nov 26, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that is correct.

    8. First I set the equation to 0 and used the quadratic formula to get that x= -0.94721 and x= -0.05279. Afterwards, I plugged it back into the original equation and got that x=22.90486 which will be absolute maximum and x= 9.541833 which will be the absolute minimum. is this correct?[/quote]
    You solved 5x2- 10x + 9= 0? That does not have real roots. And you don't want to solve that anyway. You want to set the derivative equal to 0 and solve that. Of course, max and min might occur at the endpoints of the interval.

    [quote9. f(x)= x^2 * e^8x , first i took f'(x) and got f'(x)= e^8x * 2x + x^2 * 8e^8x

    next , i took the second derivative and got f"(x)= [2x*8e^8x+2e^8x]+[8e^8x*2x + x^2*64e^8x]...

    the next step, i have to set it to zero? Can anyone help me with that and tell me if i am doing this correctly please. Thanks a lot.[/QUOTE]
    Well, what is the definition of "inflection point"?
     
  4. Nov 26, 2008 #3
    thank you for responding.

    8. f(x)= 5x^2-10x+9
    f'(x)= 10x-10

    f'(x)=0 --> 10x-10=0 --> which will give me x=1 only. how will i know the max and min given that x=1 ? I plug it back in and am i missing another root?


    9. the inflection point is when it changes from concave up to concave down or the opposite in any of the intervals
     
  5. Nov 26, 2008 #4

    Mark44

    Staff: Mentor

    Maxima/minima occur where the derivative is zero or at endpoints of the interval on which the function is defined. The graph of this function is a parabola that opens upward. It's very likely that a maximum occurs at one of the two endpoints.
    Not "it" -- the graph's concavity. This means that f''(x) is changing sign from positive to negative or from negative to positive. For this problem, find f''(x) and see where it is zero. If you find values of x for which f''(x) is zero, these values divide the real line into intervals on which f''(x) is negative or positive, which is the same as saying intervals on which the graph is concave down or concave up.
     
  6. Nov 27, 2008 #5
    8.
    f(x)=5x^2-10x+9
    f'(x)= 10x-10

    I set it to =0 and got x=1.

    Then I made a table

    [​IMG]

    which means that (0,9) is minimum and (6,129) is maximum?


    9. f"(x)= [2x*8e^8x+2e^8x]+[8e^8x*2x + x^2*64e^8x]

    I keep getting this as a second derivative and it's so long so I don't think it is right. Even if I set it to zero, how can i find out the many values of x that will make this 0? I think I know what to do next once I solve this problem. can you give me tips on a shortcut or a way how you would approach this ? thanks matt !!
     
  7. Nov 27, 2008 #6

    Mark44

    Staff: Mentor

    Ken, look at your table again. f(1) is smaller than 9, so the minimum is at (1, 9).
    Your 2nd derivative looks OK, but here it is in more readable form:
    [tex]f''(x) = 16xe^{8x} + 2e^{8x} + 16xe^{8x} + 64x^2e^{8x}[/tex]


    Factor e^(8x) out of all four terms, which leaves you with another factor that's a quadratic. Don't forget to combine like terms in the quadratic factor. e^(8x) > 0 for all real x, so the only possibility of f''(x) equaling zero comes from the quadratic being zero.

    I've checked, and there are two inflection points, both negative.
     
  8. Nov 27, 2008 #7
    8.

    for number 8, did you mean the minimum is at (0,9) or (1,9)? Then does that mean the maximum will be at (6,129)?

    9.

    I really need to learn how to do that so it won't be so confusing next time.

    [​IMG]
    Following your advice, I cancelled out the e^8x and got the quadratic equation f"(x)= 64x^2+32x+2. Then I used the quadratic formula to find where x=0 and got two inflection points just like you said.

    where x= -0.42677 and x= -0.07322

    and finally, at (-inf,C) it will be concave up . at (C,D) it will be concave down and at (D,inf) it will be concave up ?

    You should really consider being a professor. you are really good at making things clear and simple. Thanks a lot !!
     
  9. Nov 28, 2008 #8

    Mark44

    Staff: Mentor

    Sorry, I meant (1, 4) for the minimum.
    The graph will be concave up on (-inf, C) and (D, inf) if f''(x) > 0 on those intervals. Similarly, the graph will be concave down on (C, D) if f''(x) < 0 on that interval. That's pretty easy to tell because your quadratic factors into K(x + .42677)(x + .07322). Assuming K is positive, this expression is positive for x less than -.42677 and for x greater than -.07322, and the expression is negative for values of x between those two numbers. (I used your numbers, which are approximations.)
    Been there, done that. I taught math at a community college for 18 years. I guess I'm drawn to this forum because I enjoyed helping people with math, but I got tired of correcting homework and tests and such.
     
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