# Thread: domain of combined function

1. ## domain of combined function

Hi,

I hope someone can help. I'm trying to figure out the domain for the product of f(x) and g(x).

f(x) = $\log_{10}(x+3)^2$
g(x) = $\sqrt{x-1}$

I know that the domain for f(x) is x > -3 and the domain for g(x) is $x \leq -1$ or x $\geq 1$.

According to my textbook the domain for the product of f(x) and g(x) is $x \leq -1$ or x $\geq 1$, and x does not equal -3. However wouldn't the domain just be x > -3?

- Olivia

2. ## Re: domain of combined function

The domain for $g(x)$ is $x\ge 1$. Suppose $x=-1$, as you suggest. Then you are taking the square root of -2. Is that what you meant?

Next, what do you mean by the "product" of f(x) and g(x)? Do you mean
$f(x)\cdot g(x)$
or
$(f\circ g)(x)$?

3. ## Re: domain of combined function

Also, the domain for $f(x)$ is $x \neq -3$. The logarithm function can only take positive values. Anything squared is nonnegative. So, the only values not in the domain are when the input to the logarithm is zero. That only happens at $x=-3$.

4. ## Re: domain of combined function

Does $f(x) = \log_{10}(x+3)^2$ mean $[\log_{10}(x+3)]^2$ or $\log_{10}[(x+3)^2]$

5. ## Re: domain of combined function

The two functions are

And the domain of the product of the two functions, i.e. (f ∘ g)(x), according to the textbook is "all real numbers such that x is less than or equal to negative one or x is greater than or equal to one and x does not equal three". I want to know whether this domain is correct or not because I believe that it would just be "all real numbers such that x is greater than -3". Please let me know.

6. ## Re: domain of combined function

Originally Posted by otownsend
The two functions are
And the domain of the product of the two functions, i.e. (f ∘ g)(x), according to the textbook is "all real numbers such that x is less than or equal to negative one or x is greater than or equal to one and x does not equal three".
$\log(x^2+6x+9)=\log[(x+3)^2]=2\log|x+3|$. Now the domain is the set of all real numbers not equal to negative three.
Now the domain of $g$ is the set of all numbers greater than or equal to one.
There is absolutely no reason to think that $\log(x)$ is base ten.
$f\circ g(x)=2\log[\sqrt{x-1}+3]$ has domain of all real numbers greater than one.

7. ## Re: domain of combined function

Ok I'll learn to use the correct notations from now on. Thanks for your help.

8. ## Re: domain of combined function

Originally Posted by Plato
$\log(x^2+6x+9)=\log[(x+3)^2]=2\log|x+3|$. Now the domain is the set of all real numbers not equal to negative three.
Now the domain of $g$ is the set of all numbers greater than or equal to one.
There is absolutely no reason to think that $\log(x)$ is base ten.
$f\circ g(x)=2\log[\sqrt{x-1}+3]$ has domain of all real numbers greater than one.
Actually, on nearly all scientific calculators the "log" button is defined as the base 10 logarithm while the "ln" button is defined as the natural logarithm, and in many high school text books, the "log" notation IS used for the base 10.

However I agree that there does need to be much more consistency and a consensus reached. Thankfully this is the case in all higher mathematics where the "log" notation always means natural logarithm.

9. ## Re: domain of combined function

Originally Posted by otownsend
The two functions are

And the domain of the product of the two functions, i.e. (f ∘ g)(x), according to the textbook is "all real numbers such that x is less than or equal to negative one or x is greater
than or equal to one and x does not equal three". I want to know whether this domain is correct or not because I believe that it would just be "all real numbers such that x is
greater than -3". Please let me know.
The product of two functions is not (f ∘ g)(x). It is (f * g)(x), where the asterisk would be normally be a multiplication dot. So, you are not doing the
composite of two functions.

In your post # 5, you changed and clarified the functions from your original post. The answer in your textbook is correct for the product of the two functions.

For g(x), the radicand, $x^2 - 1,$ must be greater than or equal to 0. Or, $x^2 \ge 1.$ Can you see that $x \le -1$ or $x \ge 1$ for the domain of g(x)?

And, for f(x), regardless of its base, in the argument, the only restriction for its domain is that $x \ne -3$.

So, the restrictions on the two functions must be combined together for the domain of their product.