I have no idea what to do here, can anyone help?:
{$\displaystyle {(x,y)|y=2x+1}$}
I have to find the Domain, Range and Inverse.
$\displaystyle y = 2x + 1$
Domain: Are there any restrictions for x in this expression?
Range: Given the domain, are there any values that y cannot take on?
To find the inverse we must first determine that this is 1 to 1 (which it is.) Then:
$\displaystyle y = 2x + 1$
Reverse the roles of x and y. (This effectively reflects the function over the line y = x.)
$\displaystyle x = 2y + 1$
Solve for y:
$\displaystyle y = \frac{x - 1}{2}$
This is your inverse function.
-Dan
Is there any real number that makes $\displaystyle 2x + 1$ undefined or otherwise impossible to calculate? No. Thus the domain is all real numbers.
The situations you need to worry about are functions with vertical asymptotes or that are undefined. For example
$\displaystyle f(x) = \frac{g(x)}{h(x)}$
is undefined for any values of x that make h(x) = 0. Similarly we can't have any negative numbers or zero in $\displaystyle ln(x)$. We also have the case for any "even" radicals that we forbid negative numbers: $\displaystyle x^{1/2} = \sqrt{x}$, or $\displaystyle x^{1/4} = \sqrt[4]{x}$, etc. Since your function contains none of these expressions there is nothing to worry about.
-Dan
The domain is all values of x from 0 to 4. As this is a linear function the range is easy: just calculate g(x) for the end points of the interval and that is your range.
This is slightly more complicated for a non-linear function. For example, using the same domain,
$\displaystyle h(x) = \frac{1}{x - 2}$
has a range of all real numbers. The best strategy that I have is to graph the function and visually determine the range.
-Dan
Your endpoints are 0 and 4. So g(0) = ? g(4) = ?
And, of course, note that your domain is $\displaystyle (0, 4)$ which does not include either x = 0 or x = 4 so you have to make sure you write your range interval to not include those endpoints.
-Dan