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  1. #1
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    singularity

    Can someone help me with this, I don't really understand how to do it, if you help me with one step by step i could figure out the rest.

    For each real function described , give all the values of v for which the function has singularities and tell whether the singularities are removable. If the singularities are removable, redefine the function to make it continuous.
    a)f(x)=tanx

    b)g(t)=((t^2)-t)/(t-1)

    c)h(a)=a/((3a^2)-2a-8)

    d)j(y)=(y-1)(y+2)(y-3)
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  2. #2
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    Re: singularity

    Quote Originally Posted by franios View Post
    Can someone help me with this, I don't really understand how to do it, if you help me with one step by step i could figure out the rest.

    For each real function described , give all the values of v for which the function has singularities and tell whether the singularities are removable. If the singularities are removable, redefine the function to make it continuous.
    a)f(x)=tanx

    b)g(t)=((t^2)-t)/(t-1)

    c)h(a)=a/((3a^2)-2a-8)

    d)j(y)=(y-1)(y+2)(y-3)
    A sigulatiry is removable if the limit at the point exists. The function can be made continous by defining the fuction to be the value of the limit at that point.

    For example take b)

    g(t)=\frac{t^2-t}{t-1}=\frac{t(t-1)}{t-1}

    simplifing gives

    g(t)=t, \text{ if } t \ne 1

    Since the limit exists as t approches 1 we can make the function continous by "filling in" the missing value



    g(t)= \begin{cases} t, \text{ if } t \ne 1 \\1, t=1 \end{cases}

    or more simply g(t)=t

    Now try a few of the others.
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  3. #3
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    Re: singularity

    a)f(x)=tanx=sin(x)/cos(x)

    This a singularity when cos(x) = 0. cos(x)=0 when x is an odd integer multiple of pi.

    That is when x=((2n+1)pi)/2, n is any integer.

    This singularity is not removable.

    right?
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  4. #4
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    Re: singularity

    The denominator is 0 when 3a^2 - 2a - 8 = 0

    factor this: 3a^2 - 2a - 8 = 0

    3a^2 -6a + 4a - 8 = 0 3a(a - 2) + 4(a - 2) = 0

    (3a + 4)(a - 2) = 0

    a = -4/3 a = 2.

    So h(a) has singularities at a = -4/3 a = 2.

    The singularities are not removable. Another thing is that if the denominator is not 0

    at the singularities then it's definitely not removable.
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  5. #5
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    Re: singularity

    d)j(y)=(y-1)(y+2)(y-3) it has no singularity since it's a polynomial and polynomial

    are continuous for all real numbers
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