What is the domain of the function
1 answer:
Answer:
Step-by-step explanation:
The domain of a function is all of the values that can be under that specific function. In this case, we're asking what values of
allow
to exist.
In order for square roots to exist, the quantity under the square root must be greater than or equal to , because you can't take the square root of a negative number. Therefore, we can write the following inequality to solve for
:
Solving this inequality, we get:
(Subtract
from both sides of the inequality to isolate
)
(Multiply both sides of the inequality by
to get rid of
's coefficient)
Hope this helps!
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The answer is: [C]: " f(c) = 
c + 32 " .
________________________________________________________
Explanation:
________________________________________________________
Given the original function:
" c(y) = (5/9) (x <span>− 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
</span>→ <span>Write the original function as: " y = </span>(5/9) (x − 32) " ;
Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
x = (5/9) (y − 32) ;
Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
"left-hand side" of the equation:
We have:
→ x = " (
) * (y − 32) " ;
Let us simplify the "right-hand side" of the equation:
_____________________________________________________
Note the "distributive property" of multiplication:
__________________________________________
a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b – c) = ab – ac .
__________________________________________
As such:
__________________________________________
" () * (y − 32) " ;
= [ () * y ] − [ () * (32) ] ;
= [ () y ] − [ () * (
" ;
= [ () y ] − [ (
] ;
= [ () y ] − [ (
] ;
= [ (
) ] − [ ( ] ;
= [
] ;
_______________________________________________
And rewrite as:
→ " x = " ;
We want to rewrite this; solving for "y"; with "y" isolated as a "single variable" on the "left-hand side" of the equation ;
We have:
→ " x = " ;
↔ " = x ;
Multiply both sides of the equation by "9" ;
9 * = x * 9 ;
to get:
→ 5y − 160 = 9x ;
Now, add "160" to each side of the equation; as follows:
_______________________________________________________
→ 5y − 160 + 160 = 9x + 160 ;
to get:
→ 5y = 9x + 160 ;
Now, divided Each side of the equation by "5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
→ 5y / 5 = (9y + 160) / 5 ;
to get:
→ y = (9/5)x + (160/5) ;
→ y = (9/5)x + 32 ;
→ Now, remember we had substituted: "y" for "c(f)" ;
Now that we have the "equation for the inverse" ;
→ which is: " (9/5)x + 32" ;
Remember that for the original ("non-inverse" equation); "y" was used in place of "c(f)" . We have the "inverse equation"; so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as: "f(c)" .
Note that "x = c" ;
_____________________________________________________
So, the inverse function is: " f(c) = (9/5) c + 32 " .
_____________________________________________________
The answer is: " f(c) = c + 32 " ;
_____________________________________________________
→ which is:
→ Answer choice: [C]: " f(c) = c + 32 " .
_____________________________________________________
________________________________________________________
Explanation:
________________________________________________________
Given the original function:
" c(y) = (5/9) (x <span>− 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
</span>→ <span>Write the original function as: " y = </span>(5/9) (x − 32) " ;
Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
________________________________________________________
x = (5/9) (y − 32) ;
Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
"left-hand side" of the equation:
We have:
→ x = " (
Let us simplify the "right-hand side" of the equation:
_____________________________________________________
Note the "distributive property" of multiplication:
__________________________________________
a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b – c) = ab – ac .
__________________________________________
As such:
__________________________________________
" (
= [ (
= [ (
= [ (
= [ (
= [ (
= [
_______________________________________________
And rewrite as:
→ " x =
We want to rewrite this; solving for "y"; with "y" isolated as a "single variable" on the "left-hand side" of the equation ;
We have:
→ " x =
↔ "
Multiply both sides of the equation by "9" ;
9 *
to get:
→ 5y − 160 = 9x ;
Now, add "160" to each side of the equation; as follows:
_______________________________________________________
→ 5y − 160 + 160 = 9x + 160 ;
to get:
→ 5y = 9x + 160 ;
Now, divided Each side of the equation by "5" ;
to isolate "y" on one side of the equation; & to solve for "y" ;
→ 5y / 5 = (9y + 160) / 5 ;
to get:
→ y = (9/5)x + (160/5) ;
→ y = (9/5)x + 32 ;
→ Now, remember we had substituted: "y" for "c(f)" ;
Now that we have the "equation for the inverse" ;
→ which is: " (9/5)x + 32" ;
Remember that for the original ("non-inverse" equation); "y" was used in place of "c(f)" . We have the "inverse equation"; so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as: "f(c)" .
Note that "x = c" ;
_____________________________________________________
So, the inverse function is: " f(c) = (9/5) c + 32 " .
_____________________________________________________
The answer is: " f(c) =
_____________________________________________________
→ which is:
→ Answer choice: [C]: " f(c) =
_____________________________________________________
