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Splitting a function into 2 intervals

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.

Study Forum Helper

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.

OP

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?

Study Forum Helper

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.

OP

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?

Study Forum Helper

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.

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:

:lol:

Cosh is defined for all real

x

but it's not clear what the OP is supposed to be doing here. :hmmmm:

:hmmmm:

OP

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...

Study Forum Helper

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.

(edited 7 years ago)

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