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Graphing Functions of Differentiation

  1. Apr 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the intervals of increase or decrease.
    [tex] h(x)=(x+1)^{5}-5x-2 [/tex]

    2. Relevant equations



    3. The attempt at a solution

    I found the derivative to be
    \begin{align*}
    h'(x) &= 5(x+1)^{4}-5 \\
    &= 5[(x+1)^{4}-1] \\
    &= (x^{4}+4x^{3}+6x^{2}+4x+1-1) \\
    &= (x^{4}+4x^{3}+6x^{2}+4x) \\
    &= x(x^{3}+4x^{2}+6x+4) \\
    &= x[x^{2}(x+4)+2(3x+2)]
    \end{align*} That is what I got so far.
     
    Last edited by a moderator: Apr 10, 2013
  2. jcsd
  3. Apr 8, 2013 #2

    jbunniii

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    If your goal is to find out where ##h## is increasing or decreasing, then your first expression for ##h'(x)## is probably easier to work with than the rest of the stuff you wrote. For what values of ##x## is the following inequality true?
    $$5(x+1)^4 - 5 >0$$
     
  4. Apr 8, 2013 #3
    When x > 0 and when x < -2.
     
    Last edited: Apr 8, 2013
  5. Apr 8, 2013 #4

    jbunniii

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    Please explain how you got this answer.
     
  6. Apr 8, 2013 #5
    When x is greater than 0, the first part of the equation is greater than 5. Let's say you put in 1 for x and 5*2 - 5 = 5 which is greater than 0. Also, when x is less than -2, let's use -3, then -2^4 = 16 and 5*16= 80 and 80-5= 75, so x is greater than 0.
     
  7. Apr 8, 2013 #6

    jbunniii

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    It seems like you are just plugging in values for ##x##, as opposed to solving the inequality. A more careful line of argument might start as follows:

    Observe that
    $$5(x+1)^4 - 5 > 0$$
    if and only if
    $$5(x+1)^4 > 5$$
    if and only if
    $$(x+1)^4 > 1$$
    If we put ##y = x+1## then this is equivalent to
    $$y^4 > 1$$
    Since the equation ##y^4 = 1## has exactly two real solutions (##y = \pm 1##), it follows that the equation ##(x+1)^4 = 1## has exactly two real solutions (##x = 0, -2##).

    Now how can we use this to obtain the desired conclusion?
     
    Last edited: Apr 8, 2013
  8. Apr 8, 2013 #7
    The book has 0 and -2 as critical values.
     
  9. Apr 8, 2013 #8

    jbunniii

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    Thanks, I edited my previous post to fix the sign error. So what can you say about the sign of ##h'(x)## in each of the following intervals: ##(-\infty, -2)##, ##(-2, 0)##, ##(0, \infty)##?
     
  10. Apr 8, 2013 #9
    It is increasing on (-∞,-2) and (0,∞) and decreasing on (-2,0) because x is greater than 0 on the intervals (-∞,-2) and (0,∞) and less than 0 on (-2,0).
     
  11. Apr 9, 2013 #10

    jbunniii

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    Looks right to me.
     
  12. Apr 9, 2013 #11

    HallsofIvy

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    The derivative in this case is continuous. Once you have found the "critical points", where the derivative is 0, the derivative can change from "+" to "-" only where it is 0 so "plugging in values of x" in each interval is a perfectly valid way of solving the inequality.
     
  13. Apr 10, 2013 #12
    I found that to be easier, because I still couldn't find out from the derivative how -2 was a critical number.
     
  14. Apr 10, 2013 #13

    SteamKing

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    What's the value of the derivative when x = -2?
     
  15. Apr 10, 2013 #14
    The answer is 0.
     
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