# Mathematics-Functions help

1. Jul 3, 2007

### a150daysflood

Functions f and g are defined as follows:
f:x -->ln(2+x) , x> -2
g:x-->1/(x+3)+2 , x<-3

(ii)Explain why the composite function fg cannot be properly defined unless the domain of g is restricted to a subset of {where x is real,x < -3}.State the largest possible subset for fg to exist and find the corresponding range.

My problem:
Domain of f is x>-2,
Range of g is (2,-infinity].
So g is not a subset of f,then why did the question says that it can be properly defined if it is restricted to {where x is real,x < -3}?
Please enlighten me thank you.

2. Jul 3, 2007

### HallsofIvy

Staff Emeritus
?? It does't say that! It says that it "cannot be properly defined unles the domain of g is restricted to a subset of {x< -3}".

As you say, the domain of f is {x> -2}. In order that f(g(x)) be defined, we must have g(x)> -2. That is the same as saying
$$\frac{1}{x+3}+ 2> -2$$
[tex]\frac{1}{x+3}> -4[/itex]
Since the definition of g restricts x to x< -3, x+ 3< 0. Multiplying both sides of the inequality by the negative number x+ 3 we get
[tex]1< -4(x+3)= -4x- 12[/itex]
[tex]13< -4x[/itex]
$x> -13/4= -3.25$
Since x must be less than -3 by the definition of g, f(g(x)) is defined only for -3.25< x< -3.

3. Jul 3, 2007

### Dick

I think you want x<-13/4. Looks like in the last step you flipped both the direction of the inequality and the side of the inequality the x is on.

4. Jul 22, 2007

### girlsridemx2

i have a few questions that i just cant seem to understand, if you could PLEASE help me i'd appreciate it!
Consider the quadratic function f(x)=2x^2+4x-3
what is the domain and range of f(x)?

and
Find ALL asymptotes of the function f(x)=x^2-x-12
______________
x-2
* thats suppose to be over x-2

and lastly Find the x- and y- intercepts of f(x)= x^2-x-12
____________
x-2

They might be SUPER easy to you guys but im just not understanding. Thanks for the time that you've looked over this.
email me at girlsridemx2@yahoo.com

5. Jul 22, 2007

### symbolipoint

girlsridemx2,
I'll help with some of that:

For your first f(x), no restriction is needed in the use of all real numbers. You can perform completion of the square to convert into standard form and find the minimum (x, y) vertex. This will tell you your range for the function. The y values will be all real numbers greater than or equal to f(x) for this particular x value.

To help with your second example, remember that division by zero is impossible; so (x-2) will never be zero, meaning x must never equal negative 2. The function there will also have a minimum point, so the values for range will be f(x) for the vertex point and all real numbers greater than this but EXCLUDING x=2

6. Jul 23, 2007

### Schrodinger's Dog

x intercepts are f(x)=0 when f(x)= x^2-x-12=0

by factorising $x^2-x-12=0 \rightarrow (x+3)(x-4)=0$

so this happens when x=4 and x=-3

the y intercept is at f(0) or x=0, I'm sure you can work that out.

Last edited: Jul 23, 2007
7. Jul 23, 2007

### girlsridemx2

thank you so much! that make a lot more sense.