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N! asked in Science & MathematicsMathematics · 8 years ago

How do you find the domain of g(t)=ln(cos(t)) and g(t)=(sqrt(1-e^t))?

can someone explain how to find the domain of g(t)=ln(cos(t)) and g(t)=(sqrt(1-e^t))?

Thanks in advance

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  • 8 years ago
    Favorite Answer

    ===== g(t)=ln(cos(t))

    First, the domain of

    cos(t)

    is all real numbers

    the range is [ -1, +1 ]

    The domain of

    ln(x)

    is all real numbers > 0

    as ln(x) is not defined for numbers less than or equal to 0

    for g(t) = ln(cos(t))

    as the cos function is embedded in the ln function, the range of the cos function becomes the domain of the ln function

    the cos function range is [ -1, +1 ] but we have to shrink that down to

    ( 0, +1 ]

    Now we have to figure out the domain from that range, as we are no longer able to use all the real numbers.

    The cos function is a repeating function, and for its range to be (0,+1] that means the domain must be only numbers in half of that function. I could explain this better with a graph. The domain will end up being

    -90 degrees < t <= +90 degrees

    or

    -pi/2 < t <= +pi/2

    ( -pi/2, +pi/2 ]

    My answer is not complete

    "notthejake" gave a more complete answer, the entire domain for this function.

    ===== g(t)=(sqrt(1-e^t))

    the domain of

    e^t

    is all real numbers

    The domain of

    sqrt(x)

    is all positive real numbers, or all numbers greater than or equal to 0

    [ 0, infinity ]

    as the square root function is not defined for negative numbers, or all numbers less than 0

    so for (1-e^t) to be within the domain of a sqrt function,

    1-e^t must be greater than or equal to 0

    so

    1-e^t = 0

    1=e^t

    ln(1) = ln(e^t) = t

    0 = t

    for t > 0

    e^t is greater than 1

    and (1-e^t) is a negative number - can't be in a sqrt function

    for t < 0

    e^t is less than 1 - it's a fraction

    and (1-e^t) is less than 1 but greater than 0

    so this domain can be in the sqrt function

    g(t)=(sqrt(1-e^t))

    [ - infinity, 0 ]

    ===== Test each one

    I will leave that job to you, just select numbers at or near the ends of each domain, and some in the middle, and evaluate the functions on a calculator

    ===== end

  • 8 years ago

    the domain of ln u is u > 0

    domain of ln (cos t) is cos t > 0

    this is true on the right half of the circle:

    -pi/2 < t < pi/2

    but we also must add complete revolutions, so:

    -pi/2 + 2pi * n < t < pi/2 + 2pi * n (where n is an integer)

    this accounts for every incidence of the right half of the unit circle where cos > 0

    ******

    domain of sqrt(u) is u >= 0

    domain of sqrt(1 - e^t) is 1 - e^t >= 0

    1 >= e^t

    ln 1 >= t

    ln 1 = 0 => t <= 0

    domain of g(t) = sqrt(1 - e^t) is t <= 0

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