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Math Help - rate of change

  1. #1
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    Unhappy rate of change

    Q1- The distance traveled by an object at time t 0 is s = f(t) = t^2 where s is in meter (m) and t in seconds (sec). Find the instantaneous velocity of the object at t = 2sec


    Q2- Find the natural domain of the given function

    h(x) = 1/1-sinx

    Q3- What are the points of discontinuity for the function f(x)=x+5/x^2-1 ?
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  2. #2
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    Quote Originally Posted by Angel Rox View Post
    Q1- The distance traveled by an object at time t 0 is s = f(t) = t^2 where s is in meter (m) and t in seconds (sec). Find the instantaneous velocity of the object at t = 2sec


    Q2- Find the natural domain of the given function

    h(x) = 1/1-sinx

    Q3- What are the points of discontinuity for the function f(x)=x+5/x^2-1 ?

    1) f'(t)=2t plug in 2 get 4 \frac{m}{s}
    2) \sin(x) \neq 1 so x \neq 0
    3) yes, x=0 is discontinuous
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  3. #3
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    Quote Originally Posted by Erdos32212 View Post
    1) f'(t)=2t plug in 2 get 4 \frac{m}{s}
    2) \sin(x) \neq 1 so x \neq 0
    3) yes, x=0 is discontinuous
    3) Is the function f(x) = \frac{x + 5}{x^2} - 1 or f(x) = \frac{x + 5}{x^2- 1}?
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  4. #4
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    Unhappy

    I cannot understand ur way of solving Mr.Erdos32212 ....

    and for Sir.Prove It ...It is the 2ND one....

    Plz I am in a hurry...
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  5. #5
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    Quote Originally Posted by Angel Rox View Post
    I cannot understand ur way of solving Mr.Erdos32212 ....

    and for Sir.Prove It ...It is the 2ND one....

    Plz I am in a hurry...
    The function is discontinuous when the denominator is 0.

    So x^2 - 1 = 0

    x^2 = 1

    x = \pm 1

    So the function is discontinuous at x = \{-1, 1\}.
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  6. #6
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    Quote Originally Posted by Angel Rox View Post
    Q1- The distance traveled by an object at time t 0 is s = f(t) = t^2 where s is in meter (m) and t in seconds (sec). Find the instantaneous velocity of the object at t = 2sec


    Q2- Find the natural domain of the given function

    h(x) = 1/1-sinx

    Q3- What are the points of discontinuity for the function f(x)=x+5/x^2-1 ?
    2. h(x) = \frac{1}{1 - \sin{x}}.

    This is a rational function, which is continuous everywhere except where the denominator is 0.

    So it is discontinuous at

    1 - \sin{x} = 0

    \sin{x} = 1

    x = \frac{\pi}{2} + \pi n, n \in \mathbf{Z}


    So the natural domain is

    x \in \mathbf{R} \backslash \{\frac{\pi}{2} + \pi n, n \in \mathbf{Z}\}
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  7. #7
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    Thanx alot Sir. I really appreciate ur way of solving and providing all the neccesary steps.
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