Math Help - functions problem

1. functions problem

HERE THE ACTIVITY

2. Originally Posted by juanpablo
plotting the filing of

f (X) = + √x-1 and answer:

[ITEM C] Look at the domain of

f (X) = + √x-1 and find one or more relationships with the intervals of positivity, negativity and invalidity of the previous table. Properly express the conclusion.

a) When is a function: g (X) = x-1?. Write the answer algebraically.

point 2 )
What values of x, f(x)=√x-1 is a function?. Analytically justify and explain why it is the function in the range found.

note
in the point 2
when say f(x)=√x-1 i want say "cube root"

Integrative Activity:

Expressed in general terms the conclusions found in part c) (see above)
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?

3. Originally Posted by juanpablo
HERE THE ACTIVITY

OK, I see the worksheet itself is a little hard to understand the way it's written, but I'll try to answer:

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

b)

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

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

Equal to zero: $x=1$

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

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

Integrative Activity: For a function $f(x)=\sqrt{h(x)}$, the domain of $f(x)$ is described by $h(x)\geq0$.

4. Originally Posted by undefined
OK, I see the worksheet itself is a little hard to understand the way it's written, but I'll try to answer:

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

b)

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

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

Equal to zero: $x=1$

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

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

Integrative Activity: For a function $f(x)=\sqrt{h(x)}$, the domain of $f(x)$ is described by $h(x)\geq0$.

thanks you friend , you help was help to me so much