# Composite f, g, and/or h

• Jan 21st 2007, 04:53 AM
symmetry
Composite f, g, and/or h
For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

(1) P(x) = 3(sqrt{x}) + 3

(2) Q(x) = sqrt({sqrt{x} + 1}) + 1

NOTE:

For question 2, the square root of x lies within another square root.

We have a square root within a square root.

Clear?
• Jan 21st 2007, 05:15 AM
earboth
Quote:

Originally Posted by symmetry
For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

(1) P(x) = 3(sqrt{x}) + 3

(2) Q(x) = sqrt({sqrt{x} + 1}) + 1

NOTE:

For question 2, the square root of x lies within another square root.

We have a square root within a square root.

Clear?

Hello,

to (1):

P(x)=g(h(x)) and then expand the bracket.

to (2):

Q(x)= h(h(x))

Substitute a complete term instead of the x of the "outer" function.

EB
• Jan 22nd 2007, 01:15 PM
symmetry
ok
Sorry...but I don't follow what you're saying.

Can you show me by doing one of the questions?

Thanks!
• Jan 23rd 2007, 06:06 AM
earboth
Quote:

Originally Posted by symmetry
For questions 1 and 2 below, let f(x) = x^2, g(x) = 3x and
h(x) = (sqrt{x}) + 1. Express each function as a composite of f, g and/or h.

(1) P(x) = 3(sqrt{x}) + 3

(2) Q(x) = sqrt({sqrt{x} + 1}) + 1

Second attempt:

to (1)

$\displaystyle P(x)=3 \sqrt{x} +3=3 \underbrace{(\sqrt{x} +1)}_{\text{this is h}}=\underbrace{3 \cdot h(x)}_{\text{this is g with h(x) instead of x}}$. Therefore:

$\displaystyle P(x)=g(h(x))$

to (2)

$\displaystyle Q(x)=\underbrace{\sqrt{\underbrace{(\sqrt{x} +1)}_{\text{this is h}}}+1}_{\text{this is h with h instead of x}}$. Therefore:

$\displaystyle Q(x)=h(h(x))$

EB
• Jan 25th 2007, 03:14 PM
symmetry
ok
Thanks. I now know what you are talking about.