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Math Help - In-equalitiies

  1. #1
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    In-equalitiies

    Friends.

    Struggling here.
    1/t > t
    i can get to (t-1)(t+1) / t > 0 but i dont know if thats right. What do i do?

    And

    e^t >= e^-t+2


    help with both would be great
    tah
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  2. #2
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    Re: In-equalitiies

    Quote Originally Posted by tankertert View Post
    Friends.

    Struggling here.
    1/t > t
    i can get to (t-1)(t+1) / t > 0 but i dont know if thats right. What do i do?

    And

    e^t >= e^-t+2


    help with both would be great
    tah
    It should be pretty obvious that \displaystyle \begin{align*} t \neq 0 \end{align*}. You need to consider two separate cases (the first where \displaystyle \begin{align*} t < 0 \end{align*} and the second where \displaystyle \begin{align*} t > 0 \end{align*}).

    Case 1: \displaystyle \begin{align*} t < 0 \end{align*}

    \displaystyle \begin{align*} \frac{1}{t} &> t \\ 1 &< t^2 \textrm{ since } t < 0 \textrm{ and multiplying by a negative switches the inequality sign} \\ t^2 &> 1 \\ \sqrt{t^2} &> \sqrt{1} \\ |t| &> 1 \\ t < -1 \textrm{ or } t &> 1 \\ t &< -1 \textrm{ since we have originally restricted }t < 0 \end{align*}


    Case 2: \displaystyle \begin{align*} t > 0 \end{align*}

    \displaystyle \begin{align*} \frac{1}{t} &> t \\ 1 &> t^2 \\ t^2 &< 1 \\ \sqrt{t^2} &< \sqrt{1} \\ |t| &< 1 \\ -1 < t &< 1 \\ 0 < t &< 1 \textrm{ since we originally restricted } t > 0 \end{align*}

    Therefore the solution to \displaystyle \begin{align*} \frac{1}{t} > t \end{align*} is \displaystyle \begin{align*} t \in (-\infty, 1) \cup (0, 1) \end{align*}.
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  3. #3
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    Re: In-equalitiies

    I feel stupid! Thankyou!


    But what about e^t >= e^-t+2
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  4. #4
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    Re: In-equalitiies

    Quote Originally Posted by tankertert View Post
    Friends.

    Struggling here.
    1/t > t
    i can get to (t-1)(t+1) / t > 0 but i dont know if thats right. What do i do?

    And

    e^t >= e^-t+2


    help with both would be great
    tah
    \displaystyle \begin{align*} e^t &\geq e^{-t + 2} \\ \ln{\left(e^t\right)} &\geq \ln{\left(e^{-t + 2}\right)} \\ t &\geq -t + 2 \\ 2t &\geq 2 \\ t &\geq 1 \end{align*}
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  5. #5
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    Re: In-equalitiies

    Should be (1-t)(1+t)/t>0 Will be >0 if 1-t,1+t and t are all >0. This requires 0<t<1. Can you think of other possibilities?
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