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Carissa15

  • one year ago

I need to find the intervals on which the following equations have zeroes, extrema, points of inflection and intervals increasing or decreasing, as well as concave up or down. I need to show these on a graph but is there a better way to present the data (before graphing) instead of just x=1 and sub into equation, then x=2, and so on? Thanks in advance

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  1. Carissa15
    • one year ago
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    Equation 1) \[f(x) =x^2 +\frac{ 1 }{ x }\] Equation 2) \[h(x) =x^3-9x^2+27x\]

  2. zzr0ck3r
    • one year ago
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    Do you know how to take the derivative?

  3. Carissa15
    • one year ago
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    yes

  4. zzr0ck3r
    • one year ago
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    K what is the derivative of the first one?

  5. Carissa15
    • one year ago
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    2x -1x^-2 ?

  6. zzr0ck3r
    • one year ago
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    right, now set that =0 and solve for x

  7. Carissa15
    • one year ago
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    not sure how to do that for the -x^-2 part? Do you add 1^2 to RHS?

  8. Carissa15
    • one year ago
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    to give\[x=\frac{ 1 }{ 2 }\]

  9. zzr0ck3r
    • one year ago
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    woops sorry

  10. zzr0ck3r
    • one year ago
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    \(2x-\frac{1}{x^2}=0\implies 2x^3-1=0\implies x=\frac{1}{\sqrt[3]{2}}\)

  11. Carissa15
    • one year ago
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    ok

  12. Carissa15
    • one year ago
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    so this gives the zero value.

  13. zzr0ck3r
    • one year ago
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    Ok, so we have some points to look at. First we have \(0\) will give an asymptote so we look to the left of \(0\) We can put in any number to the left of \(0\) in the derivative and get a negative number so we know the original equation is decreasing on the left of \(0\). So now we at the derivative between \(0\) and \(\frac{1}{\sqrt[3]{2}}\) and get that it is positive so the original equation is also decreasin on \((0, \frac{1}{\sqrt[3]{2}})\). Do the same thing to the right of the fraction solution to get its decreasing. you with me?

  14. zzr0ck3r
    • one year ago
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    Finally we need to look at the zeros of the original equation and the right and left limits at 0.

  15. zzr0ck3r
    • one year ago
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    Tie all the together and you get the graph.

  16. Carissa15
    • one year ago
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    I am with you

  17. zzr0ck3r
    • one year ago
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    Also plug in \(\frac{1}{\sqrt[3]{2}}\) to see that it is the min for \(x>0\)

  18. Carissa15
    • one year ago
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    Ok, and we tell that the function is decreasing by comparing the values of each x=number that we are using?

  19. zzr0ck3r
    • one year ago
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    we get |dw:1439178796606:dw|

  20. zzr0ck3r
    • one year ago
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    we plug in a number like -3 into the derivative, and that gives a negative values, so the rate of change is negative, so the original function is decreasing.

  21. Carissa15
    • one year ago
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    oh right ok.

  22. zzr0ck3r
    • one year ago
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    After knowing we cant have x=0 in the original equation, and seeing the derivative only had one zero, we know we only need to look in 3 intervals \((-\infty, 0),(0, \dfrac{1}{\sqrt[3]{2}}),\) and \(((\dfrac{1}{\sqrt[3]{2}}), \infty)\)

  23. zzr0ck3r
    • one year ago
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    left limit of 0 is -infty right limit is infty

  24. zzr0ck3r
    • one year ago
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    Do the same sort of thing with the second one...

  25. zzr0ck3r
    • one year ago
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    up until calc you put in the y int and the 0's and used some other little tools to figure out what some limited graphs look like, you are doing the same thing here but with most graphs, and a much better idea of what the graph looks like.

  26. Carissa15
    • one year ago
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    so these 3 intervals then become the points of inflection? which then tell us the other important information.

  27. zzr0ck3r
    • one year ago
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    points of inflection are when it changes from decreasing from increasing, that only happens once.

  28. zzr0ck3r
    • one year ago
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    they are not intervals, they are points.

  29. zzr0ck3r
    • one year ago
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    I am to lazy to evaluate that...

  30. zzr0ck3r
    • one year ago
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    errrrr wait

  31. Carissa15
    • one year ago
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    That's ok. Thank you so much. What is the extrema?

  32. Carissa15
    • one year ago
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    is it infinity?

  33. zzr0ck3r
    • one year ago
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    I just explained extrema, sorry inflection point is when it changes concavity

  34. zzr0ck3r
    • one year ago
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    For that take the derivative of the derivative and find the zeros

  35. zzr0ck3r
    • one year ago
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    if those zeros are defined in the original function, then they are inflection points

  36. zzr0ck3r
    • one year ago
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    This I dont think has any zeros and so even though the graph is concave down for x< 0 and concave up for x>0 we never actually change concavity ....because of the asymptote

  37. Carissa15
    • one year ago
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    great thank you. This is my first time doing calculus and I am finding it difficult externally.

  38. zzr0ck3r
    • one year ago
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    I bet, we did plenty here because I thought this was not so new to you. For your question, the easiest way would be to 1) find the zeros of the derivative and notice the asymptote 2) look at the intervals on the left and right of those zeros to determine behavior to find extrema 3) do the same thing for the second derivative to find inflection points.

  39. Carissa15
    • one year ago
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    Awesome, thank you :) I seem to spend hours trying to figure things out on google or khan academy. Thank you so much for your help, I should be fine with the second equation now that I know the steps :)

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