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Increasing/Decreasing Intervals and Limits

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Given that
    f(x)= x* ((ln x)^2)
    f'(x)= ((ln x)^2) + 2 ln x
    f"(x)= (2 ln x) (1/x)+ (2/x)

    (a) find the intervals on which f(x) is increasing.
    (b) find the intervals on which f(x) is concave up.
    (c) find lim f(x) as x-> +∞ and lim f(x) as x -> 0+
    2. Relevant equations



    3. The attempt at a solution
    (a) Increasing on ({1/e^2}, 1) and (1, ∞ )
    (b) x >1
    (c) As for C.. It's not in the textbook.. and I don't know where else to look. I'm teaching myself this course, and I can't figure this out for the life of me. :confused:

    Thanks!
     
  2. jcsd
  3. Jun 13, 2013 #2

    Mark44

    Staff: Mentor

    Please show your work for the three parts. That way, we don't have to work the problem for ourselves. It's usually less work for us to spot errors than to go through the whole problem.

    BTW, I figured you were self-learning, as you have asked questions that typically come in two different calculus courses.

    It might be better to focus on the differentiation first, and then the integration after you're up to speed with differentiation.
     
  4. Jun 13, 2013 #3
    (a) (ln x)^2 + 2 ln x= 0
    ln x (ln x + 2) = 0
    x= 1 or x= 1/(e^2)
    I plugged in 0.25 and 0.75 (1/(e^2) < x < 1) and got increasing for that interval, then plugged in 2 and 3 for the x >1 interval and also got increasing. For x < 1/(e^2) I plugged in 0.05 and 0.10 and got decreasing on that interval.

    (b) x= 1 is a zero, I think: (2 ln(1)) (1/1) + (2/1)= 0 * (1 + 2)= 0 * 3= 0
    x <1 gives me negative numbers, while x > 1 gives me positive numbers. That is why I think x > 1 is concave up.

    (c) No clue how to start. At all. :/

    And yes, I'm just going by order of the textbook and trying to solve all the problems in here :P I will practice more differentiation for sure though. Thank you for your input.
     
  5. Jun 14, 2013 #4

    CAF123

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    Check this again. To find intervals of increase or decrease, look at the sign of the derivative. I think you subbed your values into f.

    You made an error in the first line there

    Perhaps think of things a little intuitively first: x grows much quicker than logx for large x. What does this hint at? For the second limit, consider 'l'Hospital.
     
  6. Jun 14, 2013 #5
    (a) I substituted the values into the first derivative, but I will redo the calculations just to make sure I didn't mess up any signs. The values of the derivative are consistently negative when 1/(e^2) < x < 1 . The values are all positive on the interval x > 1. The values on the interval x < 1/(e^2) are negative AND positive (I didn't get this earlier, so thank you for telling me to recheck!).

    (b) I'm sorry. I don't realize what error I made. Am I choosing the right derivative (second)? I rechecked my calculation and can't seem to figure out what I did wrong there.

    (c) The limit as x tends to positive infinity for x(ln x)^2 would be positive infinity since it would seem to 'increase without bound'? The first derivative is (ln x)^2 + 2ln x (as provided), and in this graph as x tends to 0+ the y value 'increases without bound' (positive infinity)?
     
  7. Jun 15, 2013 #6

    CAF123

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    Yes, that is what I get.
    Do you mean the value of the derivative is postive and negative for x < 1/e2?. How is this possible? This would indicate there existed another extrema for x < 1/e2.

    When you plug x=1 into the second derivative, it is not zero.
    Yes.
    No, apply l'Hospital on a form f/g
     
  8. Jun 15, 2013 #7
    (a) I clearly need to recheck my values on the interval for x< 1/(e^2). The derivative is always negative, now that I recheck it. Since the derivative is positive on the intervals 1/(e^2) < x < 1 and x > 1, the function is increasing on these intervals?

    (b) The reason I thought it makes the entire expression 0 is because when you plug in 1, into '2 ln x' you get 2 times 0 which is zero, which makes the entire expression 0? I'm not sure if I'm understanding the way the expression is arranged. Does it not distribute to the other two parts of the equation?

    (c) For L'Hopital's Rule, can I arrange it like:

    [(ln x)^3)/ (ln x)] + 2 ln x ?

    Thank you.
     
  9. Jun 16, 2013 #8

    CAF123

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    You said earlier (correctly) that the derivative was negative in this interval. Why did you change your mind? Typo?

    Check this again.

    The equation is f''(x) = (2lnx)(1/x) + (2/x) = (1/x)(2lnx + 2). Now sub in x=1.

    Consider f = x(lnx)2 and reexpress it in a form g(x)/h(x) and then apply l'Hospital.
     
  10. Jun 16, 2013 #9
    Oops! That was a typo- I meant negative.

    I get positive for the two values I plugged in (0.05 and 0.10).

    I didn't realize I should rearrange it like that: Now I get for x=1, y=2. So if I solve for the zero of this formula, I get:
    (1/x) (2lnx + 2)= 0
    2lnx + 2= 0
    2lnx = -2
    lnx= -1
    So I think x is somewhere between 0.10 and 1.00?

    (ln x)^2 + x(2ln x)(1/x) is the first derivative?
     
  11. Jun 16, 2013 #10

    CAF123

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    From this equation, you can obtain an exact value for x.

    No, what I mean is express the function f as a quotient so that you can then use l'Hospital.
     
  12. Jun 16, 2013 #11
    Is the equation e^x= e^(-1) meaning x=1/e? I'm not too sure if I did the e to the power x part correctly.
     
  13. Jun 16, 2013 #12

    CAF123

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    You got the right result but with the wrong reasoning. lnx = -1 ##\Rightarrow## x = 1/e, by exponentiating. How did you get on with l'Hospital?
     
  14. Jun 16, 2013 #13
    I think for L'Hospital's rule, I get this:

    f(x)= [(ln x)^3 + 2(lnx)^2]/(ln x)
    f'(x)= ln(x) (3ln(x) + 4)/x / [1/x]
    f'(x)= ln(x) (3ln(x)+ 4)

    I'm not sure if it's right though...
     
  15. Jun 16, 2013 #14

    CAF123

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    What I meant was to rewrite f in the form $$f(x) = \frac{(\ln(x))^2}{1/x}$$The limit of this as x tends to zero yields an indeterminate form so apply l'Hospital here.
     
  16. Jun 16, 2013 #15
    I thought you meant to apply it to the second derivative-I'm sorry! >_<

    Applying L'Hospital's Rule here would give me: (2lnx)/(x)/[ln x] = (2lnx/x) * [1/lnx]= 2/x
    When you plug in zero here, you would get the limit as x tends to zero is infinity. I think.
     
  17. Jun 16, 2013 #16

    CAF123

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    What is the derivative of 1/x? It is not lnx.
     
  18. Jun 16, 2013 #17
    >_< That was a stupid mistake. It's -1/x^2

    So I would get -2lnx (x).

    Plug in zero and the lim as x tends to the right side of zero is zero. Hopefully I didn't make any idiotic mistakes this time.
     
  19. Jun 16, 2013 #18

    CAF123

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    Yes, that is right, but I think it might be a good idea to write your argument out clearer. As it stands, from -2lnx(x), I presume you used l'Hospital another time and got: $$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{(\ln x)^2}{1/x} = \lim_{x \rightarrow 0^+} \frac{2 \ln x}{-1/x} = \lim_{x \rightarrow 0^+} 2x = 0$$

    At each intermediate step, we have an indeterminate form, so l'Hospitals rule is applicable.
     
  20. Jun 16, 2013 #19
    That is the logic I used, and I will make sure to write it out step by step on paper. Thank you so much CAF123. I truly appreciate it.
     
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