# Composite Functions

• Nov 17th 2007, 10:08 PM
oceanmd
Composite Functions
f(x) = $\sqrt {3x}$
g(x) = 1+x+x^2

Pleack check and correct the following composite functions

1) f(g(x)) = $\sqrt {3+3x+3x^2}$ Domain: all real numbers. Is this correct?
2) g(f(x)) = 1+ $\sqrt {3x}$+3x Domain: x greater than 0
3) f(f(x)) = f( $\sqrt {3x}$) =3^1/2 times (3x)^1/4 I do not know how to type square root of square root. Anyway, is the answer above correct? Domain: all real numbers
4) g(g(x)) = g(1+x+x^2) = 1+(1+x+x^2)+(1+x+x^2)^2 = x^4+2x^3+4x^2+3x+3 Domain: all real numbers.

Thank you very much
• Nov 17th 2007, 10:12 PM
Jhevon
Quote:

Originally Posted by oceanmd
f(x) = $\sqrt {3x}$
g(x) = 1+x+x^2

Pleack check and correct the following composite functions

1) f(g(x)) = $\sqrt {3+3x+3x^2}$ Domain: all real numbers. Is this correct?

yes

Quote:

2) g(f(x)) = 1+ $\sqrt {3x}$+3x Domain: x greater than 0
no, what if x = 0? would that be a problem?

Quote:

3) f(f(x)) = f( $\sqrt {3x}$) =3^1/2 times (3x)^1/4 I do not know how to type square root of square root. Anyway, is the answer above correct? Domain: all real numbers
no, what if x is negative?

Quote:

4) g(g(x)) = g(1+x+x^2) = 1+(1+x+x^2)+(1+x+x^2)^2 = x^4+2x^3+4x^2+3x+3 Domain: all real numbers.
yes
• Nov 17th 2007, 11:12 PM
oceanmd
Jhevon,

First of all thank you for your help.

I still do not understand why x cannot be negative in f(http://www.mathhelpforum.com/math-he...0c0d6f50-1.gif) =3^1/2 times (3x)^1/4. Let's say x=-1, then (3 times -1)^1/4 is 1/(-3)^4 =1/81, which will make it a real solution. Where am I going wrong?

Thank you
• Nov 17th 2007, 11:20 PM
Jhevon
Quote:

Originally Posted by oceanmd
Jhevon,

First of all thank you for your help.

I still do not understand why x cannot be negative in f(http://www.mathhelpforum.com/math-he...0c0d6f50-1.gif) =3^1/2 times (3x)^1/4. Let's say x=-1, then (3 times -1)^1/4 is 1/(-3)^4 =1/81, which will make it a real solution. Where am I going wrong?

Thank you

one thing you must get about composite functions is that they require the domain to work for the function being inputed.

say we have two functions $h(x)$ and $k(x)$

then $h(k(x))$ means we take the output for the function $k$ and plug it into $h$. but that means the $x$'s that we picked must be those that work in $k$, otherwise it is meaningless.

an explicit example would be this:

$f(x) = x^2$ and $g(x) = \sqrt{x}$

then $f(g(x)) = (\sqrt{x})^2 = x$

what is the domain here? $y = x$ is a function defined for all $x$, does that mean the domain here is all real $x$? No! it does not. the domain here is $x \in [0, \infty)$, because we have to make sure the domain works for $g$, since the values we can plug into $f$ comes from the output of $g$, and the outputs of $g$ do not work for all $x$
• Nov 17th 2007, 11:49 PM
oceanmd
Jhevon,

Got it. Will you please check one more.
f(x) = $\sqrt {x-3}$
g(x) = 3/x
1. f(g(x)) = f(3/x) = $\sqrt {(3-3x)/x}$
Question: how to solve (3-3x)/x greater or equal 0?
Domain: x cannot be 0 and ?

2. g(f(x)) = g( $\sqrt {x-3}$ = 3/ $\sqrt{x-3}$
domain: x greater than 3

3.f(f(x)) = f( $\sqrt {x-3}$ = ((x-3)^1/2-3)^1/2
Domain: x greater or equal 12

4. g(g(x)) =g(3/x) = x
Domain: x canot be 0

Thank you
• Nov 18th 2007, 12:42 AM
angel.white
Quote:

Originally Posted by oceanmd
I still do not understand why x cannot be negative in f(http://www.mathhelpforum.com/math-he...0c0d6f50-1.gif) =3^1/2 times (3x)^1/4. Let's say x=-1, then (3 times -1)^1/4 is 1/(-3)^4 =1/81, which will make it a real solution. Where am I going wrong?

This is a bit difficult to follow, try using the latex notation for a power, which is a^{b} which returns $a^{b}$

Anyway, the statement " $(3 *-1)^{1/4} \mbox{ is } \frac{1}{(-3)^{4}} =\frac{1}{81}$" is very wrong

here are some basic things for you:
$a^{b/c}=\sqrt[c]{a^{b}}=\sqrt[c]{a}^{b}$

$a^{-b}=\frac{1}{a^{b}}$

And combine them:
$a^{-b/c}=\frac{1}{\sqrt[c]{a^{b}}}=\frac{1}{(\sqrt[c]{a})^{b}}$

Here is a good post explaining fractional exponents http://www.mathhelpforum.com/math-help/51855-post3.html (it's even illustrated!)
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$(-3)^{1/4} = \sqrt[4]{-3}$, and what real number can you multiply by itself 4 times to get a negative number? there is none, so if you are taking the fourth root of a real number, that number cannot be negative. (note that this is only for even roots, if it were the third root, then it would be doable, because a negative number times itself 3 times will be negative)
• Nov 18th 2007, 12:43 AM
Jhevon
Quote:

Originally Posted by oceanmd
Jhevon,

Got it. Will you please check one more.
f(x) = $\sqrt {x-3}$
g(x) = 3/x
1. f(g(x)) = f(3/x) = $\sqrt {(3-3x)/x}$
Question: how to solve (3-3x)/x greater or equal 0?
Domain: x cannot be 0 and ?

recall that a fraction is positive if the numerator and denominator are the same sign. so we need to find for what x values do one or both of the following inequalities work for

$3 - 3x \ge 0$ AND $x > 0$ (since we can't have x = 0)

and/or $3 - 3x \le 0$ AND $x < 0$

find the solution that works for either or both of these systems
Quote:

2. g(f(x)) = g( $\sqrt {x-3}$ = 3/ $\sqrt{x-3}$
domain: x greater than 3

3.f(f(x)) = f( $\sqrt {x-3}$ = ((x-3)^1/2-3)^1/2
Domain: x greater or equal 12

4. g(g(x)) =g(3/x) = x
Domain: x canot be 0

Thank you
the rest seems fine
• Nov 18th 2007, 09:07 AM
oceanmd
Composite functions
Jhevon,

Domain for f(g(x)) = f(3/x) = http://www.mathhelpforum.com/math-he...8ddea4df-1.gif

x cannot be 0, and x is greater than 0 and less or equal 1

Is this correct?
Thank you.
• Nov 18th 2007, 09:15 AM
Jhevon
Quote:

Originally Posted by oceanmd
Jhevon,

Domain for f(g(x)) = f(3/x) = http://www.mathhelpforum.com/math-he...8ddea4df-1.gif

x cannot be 0, and x is greater than 0 and less or equal 1

Is this correct?
Thank you.

yes, but it is sufficient to say $0 < x \le 1$ or $x \in (0,1]$ or $\mbox{dom}(f) = (0,1]$ or something like that. the point is, saying x is greater than zero includes x not equal zero
• Nov 18th 2007, 09:22 AM
oceanmd
Composite Functions
Thank you, Jhevon