Continuity In a Function's Domain...4

Discussion in 'Calculus' started by nycmathguy, May 22, 2022.

  1. nycmathguy

    nycmathguy

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    Calculus
    Section 2.5

    Screenshot_20220516-084133_Samsung Notes.jpg

    IMG_20220521_190137.jpg

    Screenshot_20220521-184418_Samsung Internet.jpg

    I say B is continuous because there are no holes, gaps and jumps in its domain.
     
    Last edited: May 22, 2022
    nycmathguy, May 22, 2022
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  2. nycmathguy

    MathLover1

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    correct
     
    MathLover1, May 22, 2022
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  3. nycmathguy

    nycmathguy

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    Questions

    1. For problems 27-34, how is the domain found algebraically?

    For 34, find domain algebraically .

    2. For problems 27-34, how is the range found algebraically?

    For 30, find range algebraically.

    Thanks.
     
    nycmathguy, May 22, 2022
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  4. nycmathguy

    MathLover1

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    34.
    g(t)=cos^-1(e^t-1)

    domain: all real numbers except zero
    cos^-1(e^t-1)=0.......apply inverse property
    e^t-1=1
    e^t=1+1
    e^t=2.......take log
    log(e^t)=log(2)
    t*log(e)=log(2)
    t*1=log(2)
    t=log(2)
    so, domain is
    {t element R : t<=log(2)}


    (-infinity,log(2)]



    range:

    Since cos^-1 is a decreasing function with range of 0<=cos^-1(t)< pi and e^t-1>-1, then
    0<=cos^-1( e^t-1)< cos^-1(-1)
    0<=cos^-1( e^t-1)< π
    so, range is
    {g element R : 0<=g<π}

    For 30, follow steps above and find range algebraically.
     
    MathLover1, May 22, 2022
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  5. nycmathguy

    nycmathguy

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    Perfect. More notes for me to review when time allows.
     
    nycmathguy, May 22, 2022
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