anonymous
  • anonymous
if f(x)=[sqrt{1-x^2}] and g(x)=[sqrt{x}] Domain and range for (f+g)of and go(f*g)=?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
hey amistre64 hello
amistre64
  • amistre64
the domain of the other stuff still depends on the domains of the originals; see there they intersect for the overall domain
amistre64
  • amistre64
Hi hamid

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amistre64
  • amistre64
f(x); D = {x | 1-x^2 >= 0} g(x); D = {x | x >= 0}
amistre64
  • amistre64
the domain of the rest of the stuff would amount to the intersection of those two if both are used. and also would need to be amended if there is a division of one function by the other
anonymous
  • anonymous
wouldn't need to be amended.it's true
amistre64
  • amistre64
the range can then be determined by the joined pairs as they compare to the domian
amistre64
  • amistre64
so 1-x^2 >=0 1>= x^2 +- 1 >= x ; such that x=0 is true <------------------------> [-1 0 1] and x>=0 <------------------------> [0 1 ) the intersection is defined by the interval: [0,1]
amistre64
  • amistre64
do those [ ] mean anything important like a ceiling function perhaps?
amistre64
  • amistre64
i think i recall [...] being notation for the greatest integer function
anonymous
  • anonymous
i don't know
anonymous
  • anonymous
no [..] means \[\sqrt{?}\]
amistre64
  • amistre64
sqrt(...) tends to mean square root of some stuff so dies this imply that this is the sqrt of the sqrt of x?
amistre64
  • amistre64
i think you might have been attempting a psuedo latex code
amistre64
  • amistre64
\[\sqrt{1-x^2}\]
anonymous
  • anonymous
yes
amistre64
  • amistre64
... ok, then my fears are allayed :)
anonymous
  • anonymous
خیلی خری
amistre64
  • amistre64
\[f(x)=\sqrt{1-x^2}\] \[g(x)=\sqrt{x}\] Domain and range for (f+g)of and go(f*g) \[(f(x)+g(x))\ o\ f(x)=\sqrt{1-(\sqrt{1-x^2})^2}+\sqrt{(\sqrt{1-x^2})}\] \[\] hmmm, i hope im interping that correctly
anonymous
  • anonymous
yes yes :)D
amistre64
  • amistre64
\[g\ o\ (f(x)*g(x))=\sqrt{\sqrt{1-x^2}*\sqrt{x}\ }\] \[g\ o\ (f(x)*g(x))=\sqrt{\sqrt{x-x^3}\ }\] \[g\ o\ (f(x)*g(x))=\sqrt[4]{x-x^3}\]
amistre64
  • amistre64
\[(f(x)+g(x))\ o\ f(x)=\sqrt{1-1-x^2}+\sqrt[4]{1-x^2}\] \[(f(x)+g(x))\ o\ f(x)=\sqrt{-x^2}+\sqrt[4]{1-x^2}\] \[(f(x)+g(x))\ o\ f(x)=i\sqrt{x^2}+\sqrt[4]{1-x^2}\] that one might be tricky if we are to stay in the reals
amistre64
  • amistre64
so our new function have to exist within modified domains that intersect with our elementary domain
anonymous
  • anonymous
بله
amistre64
  • amistre64
x-x^3 >=0 x(1-x^2)>=0 x(1-x)(1+x)>=0 defines the zeros <-----------------------> -1 0 1 test for signage within the segments 2(1-2)(1+2)>=0 + - + fails -2(1+2)(1-2)>=0 -+- pass .5(1-.5)(1+.5)>=0 +++ pass -.5(1-.5)(1+.5)>=0 -++ fail + - + - <-----------------------> -1 0 1 well, the good news is that our elementary domain is still intact :)
anonymous
  • anonymous
well!
amistre64
  • amistre64
for the " i sqrt(x^2)" one I think the only real value we can use is 0; so our domain would be 0 and our range 1 on that one
anonymous
  • anonymous
فکر کنم شما خیلی زحمت کشیده اید
amistre64
  • amistre64
\[h(x)=\sqrt[4]{x-x^3}\] we should determine of there is a max within the interval of domain for this one by taking the derivaitve \[\frac{1-3x^2}{3} \sqrt[3]{x-x^3}=0\] \[x=\pm\frac{1}{3}\] x-x^3 = 0 x(1-x^2) = 0\[x=0,\pm1\] our interests are then at x=0,1/3, and 1 h(0) = 0 h(1) = 0 \[h(1/3) = \sqrt[4]{1/3-1/27}=\sqrt{26/27}\] this looks like the highest point to me then sooo range = [0,sqrt(26/27)]
amistre64
  • amistre64
my squiggly is a bit rusty; google translates as" you think I am too labored"
anonymous
  • anonymous
i'm from iran where are you from? thank you very much for solve
amistre64
  • amistre64
im sitting down here in florida
amistre64
  • amistre64
or it might be over here in florida, my geography aint none to good
anonymous
  • anonymous
where is florida?

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