F(g(x))

**jmanna98** · Sep 14th 2011, 07:52 PM

Can someone please confirm my answer on this question.

F(x)=sqrt(x), G(x)=cubedrt(1-x)

First I got sqrt(cubedrt(1-x)= sqrt(1-x)^1/3 =[(1-x)^1/3]^1/2= (1-x)^1/5?

**pickslides** · Sep 14th 2011, 08:05 PM

Or is it $\displaystyle \sqrt{\sqrt[3]{1-x}} = ((1-x)^{\frac{1}{3}})^\frac{1}{2}= (1-x)^{\frac{1}{6}}$ ??

**jmanna98** · Sep 14th 2011, 08:10 PM

of course ...thats what i meant

so would the domain be the domain of g and the domain of f before the combination? Also, I am just starting this calculus class and there is all these fancy symbols and ways to write the domain and range. What would you recommend is the simplest form of expressing domain and range?

**pickslides** · Sep 14th 2011, 08:24 PM

Originally Posted by jmanna98

of course ...thats what i meant

so would the domain be the domain of g and the domain of f before the combination?

Maybe not, what is the domains of f and the domain of g?

Originally Posted by jmanna98

Also, I am just starting this calculus class and there is all these fancy symbols and ways to write the domain and range. What would you recommend is the simplest form of expressing domain and range?

It like using the (,),[,] notations. Do you know them?

**jmanna98** · Sep 14th 2011, 08:30 PM

The book says fog is the set of all x in the domain of g such that g(x) is in the domain of f. In other words, (fog)(x) is defined whenever both g(x) and f(g(x)) are defined. This throws me off.

I do like the open and closed interval notation. Although, how do we specify if we are talking about domain or range?

thanks for all the help by the way

**Plato** · Sep 15th 2011, 02:47 AM

Originally Posted by jmanna98

The book says fog is the set of all x in the domain of g such that g(x) is in the domain of f. In other words, (fog)(x) is defined whenever both g(x) and f(g(x)) are defined. This throws me off.
I do like the open and closed interval notation. Although, how do we specify if we are talking about domain or range?

Because $\displaystyle \sqrt{\sqrt[3]{1-x}} = ((1-x)^{\frac{1}{3}})^\frac{1}{2}= (1-x)^{\frac{1}{6}}$
it is clear that the domain of $f\circ g$ is $x\le 1$ .
In interval notation that is $(-\infty,1]$ .

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