SOLUTION: I've been wrapping my head around this for a while and would like an easier explanation to solving these functions. Thank-you.
Suppose that the functions p and q are defined as
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Question 1123452: I've been wrapping my head around this for a while and would like an easier explanation to solving these functions. Thank-you.
Suppose that the functions p and q are defined as follows.
p(x)= x^2+9
q(x)= √x+8
Find:
(p ∘ q)(8) =
(q ∘ p)(8) =
Found 2 solutions by Boreal, solver91311:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
p o q put q into wherever there is a x. So the x^2 becomes (q's function^2) then add 9.
sqrt(x+8)^2+9=x+8+9=x+17 and when x=8, the function equals 25
q o p put p into q wherever there is a x. So the sqrt (x+8) becomes sqrt (x^2+9 then +8)
sqrt(x^2+9+8)=sqrt(x^2+17), so when x=8 sqrt (81)=9
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Let's say you have a function \ =\ \frac{x^2\ +\ 1}{x\ -\ 5})
Then
\ =\ \frac{(2)^2\ +\ 1}{2\ -\ 5})
\ =\ \frac{a^2\ +\ 1}{a\ -\ 5})
as long as 
\ =\ \frac{(TheArtistFormerlyKnownAsPrince)^2\ +\ 1}{TheArtistFormerlyKnownAsPrince\ -\ 5)
as long as 
The function (x))
is EXACTLY the same thing as saying \))
. In other words, take the definition of )
and every place you see an 
put the definition of )
in its place.
(x)\ =\ p\(q(x)\)\ =\ \(q(x)\)^2\ +\ 9\ =\ \(\sqrt{x}\ +\ 8\)^2\ +\ 9\ =\ x\ +\ 16\sqrt{x}\ +\ 64\ +\ 9\ =\ x\ +\ 16\sqrt{x}\ +\ 73)
Then for (8))
just plug in 8 and do the arithmetic. Hint: For the exact answer is simplest form use 
Do the other one the same way.
By the way, this solution is incorrect if you actually meant \ =\ \sqrt{x\ +\ 8})
instead of \ =\ \sqrt{x}\ +\ 8)
. If this is the case, it is on you because you failed to put parentheses around the radicand in the definition of 
. My interpretation is correct based on what you wrote.
John
My calculator said it, I believe it, that settles it
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