# Thread: Find the largest value of c for which the composite function gf is defined?

1. ## Find the largest value of c for which the composite function gf is defined?

f(x) = 3 - 4x - x^2 , x < c
g(x) = ln(3-x) , x < 3

I am able to find the composite function and it is

g(f(x)) = ln (3 - (3-4x-x^2)) = ln (4x+x^2)

I understand that g(f(x)) must be greater than 0 and that x cannot be 0 or -4 but how do i find the largest value of c for this composite function?

2. ## Re: Find the largest value of c for which the composite function gf is defined?

Unfortunately the one thing you say you "understand" is not true. There is NO condition here that "g(f(x)) must be greater than 0". The important point is that the "x" in ln(x) must be 0. So the "3- x" in g(x)= ln(3-x) must be greater than 0. That is: 3- x> 0 and so x< 3: x must be less than 3. Now, since, in g(f(x)), x has been replaced by f(x), we know that f(x) must be less than 3. f(x)= 3- 4x- x^2< 3 or, subtracting 3 from both sides, -4x- x^2< 0 which is the same as x^2+ 4x= x(x+ 4)> 0. A product of two numbers is positive if and only if both numbers have the same side so either ((x> 0) and (x+ 4>0)) or (x< 0 and x 4< 0).

That would give two intervals on which g(f(x)) is defined but the condition that "x< c" restricts us to one of those intervals.

3. ## Re: Find the largest value of c for which the composite function gf is defined?

Originally Posted by HallsofIvy
Unfortunately the one thing you say you "understand" is not true. There is NO condition here that "g(f(x)) must be greater than 0". The important point is that the "x" in ln(x) must be 0. So the "3- x" in g(x)= ln(3-x) must be greater than 0. That is: 3- x> 0 and so x< 3: x must be less than 3. Now, since, in g(f(x)), x has been replaced by f(x), we know that f(x) must be less than 3. f(x)= 3- 4x- x^2< 3 or, subtracting 3 from both sides, -4x- x^2< 0 which is the same as x^2+ 4x= x(x+ 4)> 0. A product of two numbers is positive if and only if both numbers have the same side so either ((x> 0) and (x+ 4>0)) or (x< 0 and x 4< 0).

That would give two intervals on which g(f(x)) is defined but the condition that "x< c" restricts us to one of those intervals.
Hi

4. ## Re: Find the largest value of c for which the composite function gf is defined?

Do you understand what a function is? It is like a machine. You give it input; it does something with that input; and it produces output. The function $\ln(x)$ can take only input where $x>0$. Now, $g(x) = \ln(3-x)$ says take the function $\ln(x)$ and give it the input $3-x$. Since $\ln(x)$ requires its input to be greater than zero, you now know $0 < 3-x$, so $x<3$. Now, $f(x) = 3-4x-x^2$ is defined over all real numbers $x$. So, it can take any input. Then, $g(f(x))$ says that you are giving $f(x)$ as input to the function $g(x)$. So, we just found that $f(x) < 3$ is the requirement for this to be valid input.

Since $f(x) = 3-4x-x^2 < 3$, as HallsofIvy stated, $0 < 4x+x^2 = x(4+x)$. Since the product of two numbers is positive if and only if they share the same sign (both are positive or both are negative), it must be that either $(x>0\text{ AND }4+x>0) \text{ OR } (x<0\text{ AND } 4+x<0)$. If $x>0\text{ AND }4+x>0$, then $x>0\text{ AND } x>-4$. Suppose $x=-5$. Is $x>0$ AND $x>-4$? No, that is false for both inequalities. Suppose $x = -2$. Then $x>-4$, but $x<0$, so it satisfies only one of the inequalities, but not both. So, for any $x>0$, it is guaranteed that $x>-4$. In other words, the two inequalities simplify to just one: $x>0$.

Similarly, the two inequalities $x < 0\text{ AND }4+x < 0$ simplify to just $x < -4$. This is because if $-4 \le x < 0$, then $x$ will not be less than $-4$, so it will only satisfy BOTH inequalities if $x < -4$.

Now, the problem asks for the largest value $c$ such that if $x, then $g(f(x))$ is defined. If $c = -4$, this is satisfied (since for all $x<-4$, $f(x)<3$, so $g(f(x))$ is defined). If $c>-4$, then for $-4 = x < c$, $f(x)= 3$, so $g(f(x))$ is not defined. Hence, the largest possible value for $c$ is $-4$.

If you still don't really understand, I am afraid you may need more help than could be provided over the internet. I am already just rewording what HallsofIvy already stated.

5. ## Re: Find the largest value of c for which the composite function gf is defined?

Originally Posted by HallsofIvy
Unfortunately the one thing you say you "understand" is not true. There is NO condition here that "g(f(x)) must be greater than 0". The important point is that the "x" in ln(x) must be 0.
This should have been 'the "x" in ln(x) must be greater than 0. That's part of the definition of "ln(x)". ln(x) is not defined if x is less than or equal to 0.

So the "3- x" in g(x)= ln(3-x) must be greater than 0. That is: 3- x> 0 and so x< 3: x must be less than 3.
Since the argument of ln must be positive, in order that ln(3- x) be defined 3- x must be larger than 0. Adding x to both sides of 3- x> 0, 3> x so x must be less than 3.

Now, since, in g(f(x)), x has been replaced by f(x), we know that f(x) must be less than 3. f(x)= 3- 4x- x^2< 3 or, subtracting 3 from both sides, -4x- x^2< 0 which is the same as x^2+ 4x= x(x+ 4)> 0. A product of two numbers is positive if and only if both numbers have the same side so either ((x> 0) and (x+ 4>0)) or (x< 0 and x 4< 0).

That would give two intervals on which g(f(x)) is defined but the condition that "x< c" restricts us to one of those intervals.
What, exactly, is confusing you?