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Sydnee Jacobs

Dec 02, 2024

MATH QUESTION please help I’m stumped :( Suppose that functions q and r are defined as follows. q(x)=x^2+5 r(x)=√x+3 find the following:?

Title: "Math Question: Help Needed with Composite Functions"

Body:
Suppose the functions ( q(x) = x^2 + 5 ) and ( r(x) = \sqrt{x} + 3 ) are defined as follows.

Determine the values of the following composite functions:
1. ( (r \circ q)(1) )
2. ( (q \circ r)(1) )

3 Answers

A
Anonymous

Feb 04, 2025

q(x) = x^2 + 5

r(x)= √x + 3

The main thig is to understand what r o q(1) and (q o r)(1) mean.

To get r o q we replace the x in r(x)= √x + 3 by q(x) = x^2 + 5

r o q = √(x^2 + 5) + 3

If we evaluate that with x = 1 we get

r o q(1) = √(1^2 + 5) + 5 = 5 + √(6)

In a similar way, but with the functions the other way round,

to get q o r we replace the x in q(x) = x^2 + 5 by r(x)= √x + 3

q o r = (√x + 3)^2 + 5 = 9 + 6√x + x + 5 = x + 6√x + 14

If we evaluate that with x = 1 we get

q o r(1) = 1 + 6 + 14 = 21

(r∘q)(x) = r(q(x))

∴ (r∘q)(x) = √q(x) + 3

∴ (r∘q)(x) = √(x² + 5) + 3

∴ (r∘q)(1) = √(1² + 5) + 3

∴ (r∘q)(1) = √6 + 3

(q∘r)(x) = q(r(x))

∴ (q∘r)(x) = (r(x))² + 5

∴ (q∘r)(x) = (√x + 3)² + 5

∴ (q∘r)(1) = (√1 + 3)² + 5

∴ (q∘r)(1) = (1 + 3)² + 5

∴ (q∘r)(1) = 4² + 5

∴ (q∘r)(1) = 16 + 5

∴ (q∘r)(1) = 21

Now just in case you meant r(x) = √(x + 3) in stead

(r∘q)(x) = r(q(x))

∴ (r∘q)(x) = √(q(x) + 3)

∴ (r∘q)(x) = √((x² + 5) + 3)

∴ (r∘q)(x) = √(x² + 5 + 3)

∴ (r∘q)(x) = √(x² + 8)

∴ (r∘q)(1) = √(1² + 8)

∴ (r∘q)(1) = √9

∴ (r∘q)(1) = 3

and

(q∘r)(x) = q(r(x))

∴ (q∘r)(x) = (r(x))² + 5

∴ (q∘r)(x) = (√(x + 3))² + 5

∴ (q∘r)(x) = x + 3 + 5

∴ (q∘r)(x) = x + 8

∴ (q∘r)(1) = 1 + 8

∴ (q∘r)(1) = 9

A
Anonymous

Jan 24, 2025

Well,

q(x)=x^2+5 r(x) = √(x+3)

therefore :

(q o r)(1) = q( r(1) )

= q( √(1+3) )

= q( √4 )

= q(2)

= 2^2 + 5

= 4 + 5

= 9

and

(r o q)(1) = r( q(1) )

= r( 1^2 + 5 )

= r(6)

= √(6+3)

= √9

= 3

et voilà, mademoiselle !! ?

hope it’ ll help !!

PS : if you want good and complete answers don’t forget to give BAs too !! ?

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