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    Consider the function h(x)=\frac{e^x+e^{-x}}{2} for -\infty < x < \infty

    How should I split this into two intervals (-\infty,0) and (0,\infty) ?

    Thanks.
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    (Original post by AishaGirl)
    Consider the function h(x)=\frac{e^x+e^{-x}}{2} for -\infty < x < \infty

    How should I split this into two intervals (-\infty,0) and (0,\infty) ?

    Thanks.
    Just say that x\not= 0 since the split interval does not account for it.
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    (Original post by RDKGames)
    Just say that x\not= 0 since the split interval does not account for it.
    Can you explain a little more? Like how do I break it down to get the two intervals?
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    (Original post by AishaGirl)
    Can you explain a little more? Like how do I break it down to get the two intervals?
    Well the function h is defined on \mathbb{R} so when it comes to the interval (-\infty,0)\cup (0,\infty) it disregards x=0 so your function get split such that h(0) is undefined.

    In essence you have h(x)=\cosh(x)=\frac{e^x+e^{-x}}{2} where x\not=0

    At least that's how I'd define h(x) across these two intervals anyway.
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    (Original post by RDKGames)
    Well the function h is defined on \mathbb{R} so when it comes to the interval (-\infty,0)\cup (0,\infty) it disregards x=0 so your function get split such that h(0) is undefined.

    In essence you have h(x)=\cosh(x)=\frac{e^x+e^{-x}}{2} where x\not=0

    At least that's how I'd define h(x) across these two intervals anyway.
    h^\prime(x)=\frac{1}{2}[e^x+e^{-x}(-1)]=\frac{1}{2}(e^x-e{^{-x}}) so h^\prime(x)=0 ?

    \frac{1}{2}(e^x-e^{-x})=0

    e^x=e^{-x}

    e^{2x}=1

    \ln(e^{2x})=\ln1=0

    x=0

    Is this correct?
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    (Original post by AishaGirl)
    h^\prime(x)=\frac{1}{2}[e^x+e^{-x}(-1)]=\frac{1}{2}(e^x-e{^{-x}}) so h^\prime(x)=0 ?

    \frac{1}{2}(e^x-e^{-x})=0

    e^x=e^{-x}

    e^{2x}=1

    \ln(e^{2x})=\ln1=0

    x=0

    Is this correct?
    I'm not sure what you are trying to do here? Yes h'(x)=\frac{1}{2}(e^x-e^{-x}) and the stationary point is indeed at x=0 but if that's following from the previous part then we have a problem.
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    (Original post by RDKGames)
    I'm not sure what you are trying to do here? Yes h'(x)=\frac{1}{2}(e^x-e^{-x}) and the stationary point is indeed at x=0 but if that's following from the previous part then we have a problem.
    Yes this is a tad confusing.:lol:

    Cosh is defined for all real x but it's not clear what the OP is supposed to be doing here. :hmmmm:
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    (Original post by RDKGames)
    I'm not sure what you are trying to do here? Yes h'(x)=\frac{1}{2}(e^x-e^{-x}) and the stationary point is indeed at x=0 but if that's following from the previous part then we have a problem.
    I'm trying to find the interval on which h(x) is increasing. Now that I've split it into two intervals I can show that h(x) is always concave up? I think...
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    (Original post by AishaGirl)
    I'm trying to find the interval on which h(x) is increasing. Now that I've split it into two intervals I can show that h(x) is always concave up? I think...
    Well if you're trying to find when it is increasing, setting h'(x)=0 won't help with that as this is the stationary point when the function is neither increasing nor decreasing.

    If a function is increasing on an interval then h'(x)>0 on that interval.
 
 
 
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