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Mathematics 21 Online
OpenStudy (fanduekisses):

*****WHY U NO HELP ME??!!

OpenStudy (fanduekisses):

\[f(x)= \frac{ 1 }{ x^4 -1}\] and \[g(x) = \sqrt[6]{x}\]

OpenStudy (fanduekisses):

\[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\]

OpenStudy (fanduekisses):

The radical and exponent part is confusing, how will I go about that?

OpenStudy (fanduekisses):

@ganeshie8

Nnesha (nnesha):

you should convert radical to an exponent form

Nnesha (nnesha):

\[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\]index become the denomiantor of the fraction

Nnesha (nnesha):

denominator *

Nnesha (nnesha):

\(\color{blue}{\text{Originally Posted by}}\) @Fanduekisses \[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\] \(\color{blue}{\text{End of Quote}}\) btw, -1 should be in the denominator

Nnesha (nnesha):

\[\huge\rm \frac{ 1 }{(\color{ReD}{ \sqrt[6]{x}} )^4} =\frac{ 1 }{ (\color{Red}{x^{??}})^4 }\]

OpenStudy (fanduekisses):

\[\frac{ 1 }{ (x ^{1/6} )^4}\]

Nnesha (nnesha):

yes right now you can multiply 1/6 by 4 exponent rule \[\huge\rm (a^m)^x = a^{m \times x}\]

OpenStudy (fanduekisses):

\[\frac{ 1 }{ x ^{2/3} }\]

Nnesha (nnesha):

don't forget the negative one

OpenStudy (fanduekisses):

ok so the domain will be \[x \ge0 \] and

OpenStudy (fanduekisses):

Wait how do I solve for x? \[x ^{2/3}= 1\]

OpenStudy (fanduekisses):

The denominator part, to find the domain, what x cannot equal.

Nnesha (nnesha):

ugh sorry my laptop is acting up,slow net connection alright so whats the question ? find domain ?

OpenStudy (fanduekisses):

yes of the composition f(g(x))

Nnesha (nnesha):

you can take cube both sides to get rid of the fraction

Nnesha (nnesha):

like we rewrite x^2/3 into radical form \[\huge\rm \sqrt[3]{x^2}=1\] to cancel out cube root we should take cube sides just like we take square root to cancel out square

OpenStudy (fanduekisses):

wait be right back

Nnesha (nnesha):

btw, you can apply difference of squres....

OpenStudy (mathmate):

\(\frac{ 1 }{ (\sqrt[6]{x})^4 }-1=\frac{1}{x^{2/3}}-1=x^{-2/3}-1=0\) It doesn't change the answer in this case. lol

OpenStudy (fanduekisses):

So th\[x \neq1 ?\]e domain of f(g(x)) is \[x \ge0\] and

OpenStudy (fanduekisses):

Sorry for the late reply, I had an urgency.

Nnesha (nnesha):

how many solutions would you get when u take square root of something ? and it's okay :=)

OpenStudy (fanduekisses):

+- 1 hehe

OpenStudy (fanduekisses):

:D got it now hehe thanks so much!

OpenStudy (fanduekisses):

So the domain is x cannot equal +-1 and x has to be greater than or equal to 0 ?

Nnesha (nnesha):

to write in interval u can draw a number line |dw:1443037013961:dw|

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