HERE THE ACTIVITY

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- Apr 7th 2010, 12:25 PMjuanpablofunctions problem
HERE THE ACTIVITY

http://i44.tinypic.com/24enuoz.jpg - Apr 7th 2010, 01:32 PMundefined
I'd like to help, but I don't think you copied the questions down well enough for people to understand what needs to be found.

For example, there's a reference to a table, but you didn't copy the table. Also, it's not entirely clear whether f(x)=(x-1)^(1/2) or f(x)=x^(1/2)-1. And then in point 2 you redefine f(x) using cube roots, but normally a textbook or worksheet would use a different letter to avoid confusion, since we already used f(x) to mean something else.

Also the question "when will g(x)=x-1 be a function" doesn't seem to make much sense, since that is always a function, for all real x.

Would it be possible to take a picture of the problem and upload using, say, tinypic.com? - Apr 7th 2010, 02:39 PMundefined
OK, I see the worksheet itself is a little hard to understand the way it's written, but I'll try to answer:

a) $\displaystyle g(x)=x-1$ is a function for all real x. This could be written in several different ways, including $\displaystyle x\in\mathbb{R}$ and $\displaystyle x\in(-\infty, \infty)$.

b)

Greater than zero: $\displaystyle x\in(1, \infty)$

Less than zero: $\displaystyle x\in(-\infty, 1)$

Equal to zero: $\displaystyle x=1$

c) The domain of $\displaystyle f(x)$ is $\displaystyle x\in[1, \infty)$. This corresponds to $\displaystyle g(x)\geq0$.

Follow up question: $\displaystyle f(x)=\sqrt[3]{x-1}$ is a function for $\displaystyle x\in\mathbb{R}$ because all real numbers have exactly one real cube root. Thus as long as $\displaystyle x-1$ is a real number, $\displaystyle f(x)$ will associate it with a unique value. The range is also all real numbers.

Integrative Activity: For a function $\displaystyle f(x)=\sqrt{h(x)}$, the domain of $\displaystyle f(x)$ is described by $\displaystyle h(x)\geq0$. - Apr 7th 2010, 03:11 PMjuanpablo