*****WHY U NO HELP ME??!!
\[f(x)= \frac{ 1 }{ x^4 -1}\] and \[g(x) = \sqrt[6]{x}\]
\[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\]
The radical and exponent part is confusing, how will I go about that?
@ganeshie8
you should convert radical to an exponent form
\[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\]index become the denomiantor of the fraction
denominator *
\(\color{blue}{\text{Originally Posted by}}\) @Fanduekisses \[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\] \(\color{blue}{\text{End of Quote}}\) btw, -1 should be in the denominator
\[\huge\rm \frac{ 1 }{(\color{ReD}{ \sqrt[6]{x}} )^4} =\frac{ 1 }{ (\color{Red}{x^{??}})^4 }\]
\[\frac{ 1 }{ (x ^{1/6} )^4}\]
yes right now you can multiply 1/6 by 4 exponent rule \[\huge\rm (a^m)^x = a^{m \times x}\]
\[\frac{ 1 }{ x ^{2/3} }\]
don't forget the negative one
ok so the domain will be \[x \ge0 \] and
Wait how do I solve for x? \[x ^{2/3}= 1\]
The denominator part, to find the domain, what x cannot equal.
ugh sorry my laptop is acting up,slow net connection alright so whats the question ? find domain ?
yes of the composition f(g(x))
you can take cube both sides to get rid of the fraction
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
wait be right back
btw, you can apply difference of squres....
\(\frac{ 1 }{ (\sqrt[6]{x})^4 }-1=\frac{1}{x^{2/3}}-1=x^{-2/3}-1=0\) It doesn't change the answer in this case. lol
So th\[x \neq1 ?\]e domain of f(g(x)) is \[x \ge0\] and
Sorry for the late reply, I had an urgency.
how many solutions would you get when u take square root of something ? and it's okay :=)
+- 1 hehe
:D got it now hehe thanks so much!
So the domain is x cannot equal +-1 and x has to be greater than or equal to 0 ?
to write in interval u can draw a number line |dw:1443037013961:dw|
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