# a lot of confusion about Functions!!

• Oct 22nd 2013, 10:16 AM
ronanbrowne88
a lot of confusion about Functions!!
consider the following functions

Attachment 29543

I have attached a screen shot of my problem as that was easier than typing it out.

Apologies if this is posted in the wrong section, i have to do this problem by Thursday and to be honest i don't really have a clue but its important i get this right. anyway ill take a stab at it and hopefully some more experienced people can set me right.

so for the first part with the graph i need to get f(x) = g(x)
which would look something like this

f(x)=x2-1
g(x)=x+1

f(x) = g(x)
x2-1=x+1
(x+1)-(x2-1)=o
(x+1)-x2+1=0
-x2+x+2=0

S0 from here i need to get some co-ordinates from the above equation so i can use to draw a line on the graph?? i think, im not really sure where to go from here....and my maths may even be way off here like i said im unsure

moving on to next section
compute the following values:
f(2)
g(0)
f(2/3)

f(2)=(2)2-1 = 4-1=3
g(0)=(0) +1=1
f(2 over 3)= (2 over 3)2+1= ??

i think the first two may be right not sure what to do with fraction in last one.

as for the last section

f dot g (x) = f(x+1)=(x+1)2-1
g dot f (x) = g(x2-1) = (x2-1)+1
f dot (x) = f(x2-1) = (x2-1)2-1
g dot g(x) = g(x+1)=(x+1) + 1

again i think im right here? could some one confirm if i am or not

I realize this is a long post i just want to thank anyone in advance who can lend me their expertise! (Happy)
• Oct 22nd 2013, 10:39 AM
SlipEternal
Re: a lot of confusion about Functions!!
Quote:

Originally Posted by ronanbrowne88
so for the first part with the graph i need to get f(x) = g(x)
which would look something like this

f(x)=x2-1
g(x)=x+1

f(x) = g(x)
x2-1=x+1
(x+1)-(x2-1)=o
(x+1)-x2+1=0
-x2+x+2=0

This is correct so far. I would recommend multiplying both sides of the equation by -1 and then trying to factor the polynomial.

Quote:

Originally Posted by ronanbrowne88
S0 from here i need to get some co-ordinates from the above equation so i can use to draw a line on the graph?? i think, im not really sure where to go from here....and my maths may even be way off here like i said im unsure

That is correct. Making a table of values might help.

Quote:

Originally Posted by ronanbrowne88
moving on to next section
compute the following values:
f(2)
g(0)
f(2/3)

f(2)=(2)2-1 = 4-1=3
g(0)=(0) +1=1
f(2 over 3)= (2 over 3)2+1= ??

You are correct for the first two. For the last one, an exponent of 2 means multiply the number by itself. Do you remember how to multiply two fractions? I will set it up for you:

$\displaystyle f\left( \dfrac{2}{3}\right) = \left(\dfrac{2}{3}\right)^2+1 = \dfrac{2}{3}\cdot \dfrac{2}{3} + 1$

Quote:

Originally Posted by ronanbrowne88
as for the last section ill i don't know how to even attempt it if anyone could show me how to do one as a example i can try the rest.

I realize this is a long post i just want to thank anyone in advance who can lend me their expertise! (Happy)

The last section is composition of functions. So, $\displaystyle (f\circ g)(x)$ means take the function $\displaystyle f(x)$ and wherever you see an $\displaystyle x$, replace it with $\displaystyle g(x)$. So, $\displaystyle (f\circ g)(x) = f(g(x)) = (g(x))^2+1$. Now, in place of $\displaystyle g(x)$, put the formula for $\displaystyle g(x)$.
• Oct 22nd 2013, 11:05 AM
ronanbrowne88
Re: a lot of confusion about Functions!!

f(x)=x2-1
g(x)=x+1

f(x) = g(x)
x2-1=x+1
(x+1)-(x2-1)=o
(x+1)-x2+1=0
-x2+x+2=0
-x2+x+2=0 (next multiply by -1)

x2 -x -2 =0 (next factorize)

x2−x−2=1(x−(−1))(x−2)
x2−x−2=(x+1)(x−2)
x1=−1
x2=2

then i can use these point to draw the graph. correct??

the last one on part 2

f(2 over 3)= (2 over 3)2+1= 4 over 9 +1 = 1 and 4 over 9 (sorry cant see a button to write in correct format)

and last bit

f dot g (x) = f(x+1)=(x+1)2-1
g dot f (x) = g(x2-1) = (x2-1)+1
f dot (x) = f(x2-1) = (x2-1)2-1
g dot g(x) = g(x+1)=(x+1) + 1
• Oct 22nd 2013, 11:13 AM
SlipEternal
Re: a lot of confusion about Functions!!
Quote:

Originally Posted by ronanbrowne88
f(x)=x2-1
g(x)=x+1

f(x) = g(x)
x2-1=x+1
(x+1)-(x2-1)=o
(x+1)-x2+1=0
-x2+x+2=0
-x2+x+2=0 (next multiply by -1)

x2 -x -2 =0 (next factorize)

x2−x−2=1(x−(−1))(x−2)
x2−x−2=(x+1)(x−2)
x1=−1
x2=2

then i can use these point to draw the graph. correct??

Those points only tell you the x-coordinate where the two functions intersect. To graph the functions, you will want a table with more values for each function.

Quote:

Originally Posted by ronanbrowne88
the last one on part 2

f(2 over 3)= (2 over 3)2+1= 4 over 9 +1 = 1 and 4 over 9 (sorry cant see a button to write in correct format)

Seems we wrote down the wrong formula for $\displaystyle f(x)$. At the top, you have $\displaystyle f(x) = x^2-1$, so it should be $\displaystyle \dfrac{4}{9} - 1$. Find a common denominator, and you can simplify.

Quote:

Originally Posted by ronanbrowne88
and last bit

f dot g (x) = f(x+1)=(x+1)2-1
g dot f (x) = g(x2-1) = (x2-1)+1

Good, but you can simplify $\displaystyle (g\circ f)(x)$ (-1+1 = 0).

Quote:

Originally Posted by ronanbrowne88
f dot (x) = f(x2-1) = (x2-1)2-1
g dot g(x) = g(x+1)=(x+1) + 1

You can also simplify $\displaystyle (g\circ g)(x)$ (1+1 = 2).

Other than that, everything looks great.
• Oct 22nd 2013, 11:15 AM
ronanbrowne88
Re: a lot of confusion about Functions!!
great thanks!