Question

if f(x)=[sqrt{1-x^2}] and g(x)=[sqrt{x}]
Domain and range for (f+g)of and go(f*g)=?

Answer 1

hey amistre64 hello

Answer 2

the domain of the other stuff still depends on the domains of the originals; see there they intersect for the overall domain

Answer 3

Hi hamid

Answer 4

f(x); D = {x | 1-x^2 >= 0}
g(x); D = {x | x >= 0}

Answer 5

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

Answer 6

wouldn't need to be amended.it's true

Answer 7

the range can then be determined by the joined pairs as they compare to the domian

Answer 8

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]

Answer 9

do those [ ] mean anything important like a ceiling function perhaps?

Answer 10

i think i recall [...] being notation for the greatest integer function

Answer 11

i don't know

Answer 12

no [..] means \[\sqrt{?}\]

Answer 13

sqrt(...) tends to mean square root of some stuff

so dies this imply that this is the sqrt of the sqrt of x?

Answer 14

i think you might have been attempting a psuedo latex code

Answer 15

\[\sqrt{1-x^2}\]

Answer 16

yes

Answer 17

... ok, then my fears are allayed :)

Answer 18

خیلی خری

Answer 19

\[f(x)=\sqrt{1-x^2}\] \[g(x)=\sqrt{x}\]

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

Answer 20

yes yes :)D

Answer 21

\[g\ o\ (f(x)*g(x))=\sqrt{\sqrt{1-x^2}*\sqrt{x}\ }\]
\[g\ o\ (f(x)*g(x))=\sqrt{\sqrt{x-x^3}\ }\]
\[g\ o\ (f(x)*g(x))=\sqrt[4]{x-x^3}\]

Answer 22

\[(f(x)+g(x))\ o\ f(x)=\sqrt{1-1-x^2}+\sqrt[4]{1-x^2}\]
\[(f(x)+g(x))\ o\ f(x)=\sqrt{-x^2}+\sqrt[4]{1-x^2}\]
\[(f(x)+g(x))\ o\ f(x)=i\sqrt{x^2}+\sqrt[4]{1-x^2}\]
that one might be tricky if we are to stay in the reals

Answer 23

so our new function have to exist within modified domains that intersect with our elementary domain

Answer 24

بله

Answer 25

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 :)

Answer 26

well!

Answer 27

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

Answer 28

فکر کنم شما خیلی زحمت کشیده اید

Answer 29

\[h(x)=\sqrt[4]{x-x^3}\]

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)]

Answer 30

my squiggly is a bit rusty; google translates as" you think I am too labored"

Answer 31

i'm from iran where are you from?
thank you very much for solve

Answer 32

im sitting down here in florida

Answer 33

or it might be over here in florida, my geography aint none to good

Answer 34

where is florida?