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  1. #1
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    Question first derivative test



    Hi everyone,


    On that question I found the stationary points, but I don't know how to do the first derivative test. I have the formula but my function give me headache to work with can someone help me.




    Find any stationary points of the function g(x) defined in part (a)(ii), and use the First derivative Test to classify each stationary point as a local maximum or a local minimum of g(x).


    g(x)= ex/7
    square root 10+x^2




    This equation has solutions x= 5 and x= 2. Thus the stationery points of f are at x= 5 and x= 2.

    To classify the stationary point at x= 5, we choose test points xl= -1 and xr= 1, say. Then we have
    f(xl) = f(-1)=



    Thanks again

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  2. #2
    Super Member 11rdc11's Avatar
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    ok to use the first derivative test you test between the interval

    x < 2,
    2 < x <5
    5<x

    Now plug in values and test, if you go from - to + you have a minimum and if you go from + to - you have a maximum.
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  3. #3
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    Re: first derivative test

    f'(xl)=f'(-1)=2(-1)(-2)=4>0
    f'(xr)=f'(1)=2(1)(-1)=-2<0

    so f has a local maximum at x=0, by the first derivative test.

    And I do the same with 5, am I close
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  4. #4
    Super Member 11rdc11's Avatar
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    When x = 2 it is a max, when x = 5 it is a min. Your critical points is where you may have maxs or mins. Just wondering where did you get 0 being a critical point when you said 2 and 5 were the critical points?
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  5. #5
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    Re: first derivative test

    Sorry It was suppose to be x=2 instead of x=0.
    I don't know, when I do the same with x=5 it give me 25 and -5 so that's wrong it give 2 maximum.

    I can't see how to calculate it
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  6. #6
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    Re: first derivative test

    Maybe I need to calculate with my derivative: (10+x2)-1/2x

    I should mention that I'm really bad in differentiation and I'm completely confused on what to do
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  7. #7
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    Ok so the power went out due to hurricane IKE so im going to try to answer this the best i can on my cellphone lol. Plug in x = 3 and x = 6 into the 1st derivative and see what values it gives you. Go ahead and try x = 0 too to test all the intervals.
    Last edited by 11rdc11; September 12th 2008 at 04:10 AM.
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  8. #8
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    Smile Re: first derivative test

    Thanks for helping me

    x=3, x=6 xl= -1; xr= 1

    f'(xl)= f'(-1)= 3(-1)(-3)= 9>0 x=3, local maximum
    f'(xr)= f'(1)= 3(1)(-1)= -3<0

    f'(xl)= f'(-1)= 6(-1)(-6)= 36>0 x=6,
    f'(xr)= f'(1)= 6(1)(-1)= -6<0
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  9. #9
    Super Member 11rdc11's Avatar
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    Sorry if this is a glitch in my phone but what is xl and xr? I dont have my ti89 right now so i need you to do me a favor. Find the values of these points f'(0), f'(3),f'(6) please.
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  10. #10
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    ok i figured out what you mean by xl and xr. using your method we need to test the critical points 2 and 5. so to test for 2 xl would be 0 and and xr would b 3. to test 5 xl would b 3 and xr would be 7. you will notice that when you test for 2 xl will be positive and xr will be negative. when u test for 5 xl will be negative and xr will be positive. so 2 a max and 5 a min. hopes this helps. ill write and explain it more clearly when i get power back. it just a tad difficult to type this on a cellphone
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  11. #11
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    re: first derivative

    Yes on my book, xl and xr are testing points

    f'(xl)= f'(0)= 2(0)(-2)=
    f'(xr)= f'(3)= 2(3)(0)=

    f'(xl)= f'(3)= 5(3)(-5)= -75 local min
    f'(xr)= f'(7)= 5(7)(3)= 105

    I get 0 for x=2?
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  12. #12
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    re: first derivative

    I think I got it, can I change my test points at xl= -1, xr= 1 for x=2, and for x=5, xl=2 and xr=3, which x=5 at local minimum -50 and x=2 at local maximum 4??
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  13. #13
    Super Member 11rdc11's Avatar
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    nope you cant change your xl and xr to that, the correct answer is 2 is a max and 5 a min. ill explain why when i get electricity, sorry for making you wait.
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  14. #14
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    Thumbs up re: first derivative

    No that's ok I wait, I'm already glad you helped me that far
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  15. #15
    Super Member 11rdc11's Avatar
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    Ok time to finish this up lol

    g(x) = \frac{e^{\frac{x}{7}}}{\sqrt{10+x^2}}

    you found the derivative

    g'(x) = \frac{(x^2-7x+10)e^{\frac{x}{7}}}{7(x^2+10)^{\frac{3}{2}}}

    Now set the derivative to 0 to find critical points

    g'(x) = 0

    (x^2-7x+10) = 0

    so the critical points are

    x~=~2,~x~=~5

    Now using the 1st derivative test we test for our intervals

    x~<~2,~2~<~x~<~5,~5~<~x



    ok now we try a x value for each of our intervals

    so lets try 0 for x < 2

    g'(0) = .04518

    try 3 for 2 < x < 5

    g'(3) = -.0053

    try 6 for 5 < x

    g'(6) = .00432

    Ok from this you can see that the XL side of 2 is positive and the XR side is negative making x = 2 a max. You can also see that Xl side of 5 is negative and the XR side is positive making x = 5 a min. You have to remember to stay within the intervals when picking your XL and XR. For example for the critical point 5, I could not chose XL to be 0 because it is not in the interval of 2 < x < 5

    Hope this makes sense
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