Function: g(x) = (x+4)/sqrt[7x-5); find 4g(3)+2[g(2)]^2  TOPIC_SOLVED

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Function: g(x) = (x+4)/sqrt[7x-5); find 4g(3)+2[g(2)]^2  TOPIC_SOLVED

Postby EdSewersPark on Tue May 24, 2011 11:29 pm

Let g be the function given by

Find and simplify

Should I plug in first, then go back and plug in ? I'm not sure I even know exactly what is even saying.
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Postby stapel_eliz on Wed May 25, 2011 12:56 am

EdSewersPark wrote:Let g be the function given by

Find and simplify

Should I plug in first, then go back and plug in ? I'm not sure I even know exactly what is even saying.

Since the two terms have different x-values, you'll need to evaluate numerically first; then you can combine the results.

To learn about arithmetic (such as squaring, etc) with functions, try here. :wink:
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Re: Function: g(x) = (x+4)/sqrt[7x-5); find 4g(3)+2[g(2)]^2

Postby EdSewersPark on Thu May 26, 2011 8:49 pm

Is this how to plug it in?

=
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Postby stapel_eliz on Thu May 26, 2011 8:59 pm

EdSewersPark wrote:Is this how to plug it in?

=

Yes, but you're plugging "3" into "g(x)", so the equality is properly "g(3)=...". :wink:
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Re: Function: g(x) = (x+4)/sqrt[7x-5); find 4g(3)+2[g(2)]^2

Postby EdSewersPark on Thu May 26, 2011 10:07 pm

Just g(3)? Not 4g(3)?
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Postby stapel_eliz on Thu May 26, 2011 11:32 pm

EdSewersPark wrote:Just g(3)? Not 4g(3)?

I'd missed the extra 4's you'd put in there. But note that "4*g(3)" means "four multiplied by the value of g(3)", not "g(4*3)" or whatever you've done. :wink:
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Re: Function: g(x) = (x+4)/sqrt[7x-5); find 4g(3)+2[g(2)]^2

Postby EdSewersPark on Fri May 27, 2011 12:32 am

Ah, so After plugging in g(3) I take that sum and multiply it by four?
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Postby stapel_eliz on Fri May 27, 2011 11:59 am

EdSewersPark wrote:Ah, so After plugging in g(3) I take that sum and multiply it by four?

Exactly: after evaluating g(3), multiply the result by 4. :wink:
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Re: Function: g(x) = (x+4)/sqrt[7x-5); find 4g(3)+2[g(2)]^2

Postby EdSewersPark on Fri May 27, 2011 3:57 pm

Wow, that is so much easier than I thought.
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