Slope-intercept, (determine, evaluate, domain) function, diff. quotient

[Eshmael]

Domain Functions:

f(x) = 4sq.root(1-x2), the 4 is on the outside

f(s) = sq.root(s -1)/s-4

f(x) = x-4/sq.root x

I have absolutely no clue how to go about doing any of these, I take notes but I cannot piece it together; I have never felt so helpless at anything. This isn't homework I just want to know how to be able to do slope-intercept, determine a function, evaluate it, domain functions and difference quotients becaue I have a quiz on it tomorrow at 8:30 and I need to know how to at least start to solve them.

 

[symbolipoint]

You asked for the domains of these functions according to what you expressed in pure text form:

f(x) = 4\sqrt{1-2x}

f(x) = \frac{\sqrt{s - 1}}{s} - 4

f(x) = x -\frac{4}{\sqrt{x}}

Are those the functions you meant?

 

[Eshmael]

You asked for the domains of these functions according to what you expressed in pure text form:

f(x) = 4\sqrt{1-2x}

f(x) = \frac{\sqrt{s - 1}}{s} - 4

f(x) = x -\frac{4}{\sqrt{x}}

Are those the functions you meant?

Yeah, if you could just like direct to a website which explains these as a lesson or something, that would be great.

 

[HallsofIvy]

The "domain" of a function, if nothing but the formula is given, is all values of x for which the formula can be evaluated. For most functions the only problems are:
1) You can't divide by 0
2) You can't take the square root of a negative number.

Look at the last one:
f(x)= x- \frac{4}{\sqrt{x}}
That first x is no problem- you could replace it with any number. But the other x is inside a square root- it can't be negative. Also it is in the denominator: \sqrt{0} certainly exists, it is 0. But then you would be dividing by 0 so x also cannot be 0. In stating what the domain is you would reverse that: x cannot be negative or 0 so x can be any positive number. The domain is the set of all positive real numbers.