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

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

Let g be the function given by $g(x)\,=\,\frac{x+4}{\sqrt{7x-5}}$

Find and simplify $4g(3)+2[{g(2)]^2$

Should I plug in $4g(3)$ first, then go back and plug in $2[g(2)]^2$? I'm not sure I even know exactly what $2[g(2)]^2$ is even saying.

stapel_eliz
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EdSewersPark wrote:Let g be the function given by $g(x)\,=\,\frac{x+4}{\sqrt{7x-5}}$

Find and simplify $4g(3)+2[{g(2)]^2$

Should I plug in $4g(3)$ first, then go back and plug in $2[g(2)]^2$? I'm not sure I even know exactly what $2[g(2)]^2$ 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.

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

Is this how to plug it in?

$g(x)=\frac{4[(3+4)]}{\sqrt{4[7(3)-5)]}$=

stapel_eliz
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EdSewersPark wrote:Is this how to plug it in?

$g(x)=\frac{4[(3+4)]}{\sqrt{4[7(3)-5)]}$=

Yes, but you're plugging "3" into "g(x)", so the equality is properly "g(3)=...".

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

Just g(3)? Not 4g(3)?

stapel_eliz
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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.

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

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

stapel_eliz
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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.

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

Wow, that is so much easier than I thought.