Hi Victoria,

We'll do your problems but let's look at a different function first. I want to look at g(x) = √x. The reason I want to look at this function is that it's a button on your calculator. To make it work you enter a number and then press the √ button. Sometimes it works and sometimes it doesn't. It works if the number you entered is positive or zero and it doesn't work if the number you entered is negative. The domain of g(x) is all the possible inputs x for which g(x) works. Thus the domain of g(x) = √x is all x ≥ 0. The range of g(x) is all possible outputs of the function. So what do you get as outputs of g(x). Certainly you get zero since g(0) = √0 = 0. What you don't get are negative numbers. (Don't be confused by some older textbooks that say for example that √4 = ± 2, that's not correct, √4 = 2.) But how do you know that every positive number y is √x for some x? Just let x = y² then √x = √(y²) = y. Thus every number y ≥ 0 is a square root so the range of g(x) = √x is all y ≥0.

Now what about your function y = 2x + 1? What inputs x make this function work? You can let x be any real number and calculate 2x + 1, so the domain of y = 2x + 1 ia all real numbers. What about the range? Is every real number y an output of y = 2x + 1? Suppose I give you a number y, say y = 7, and ask you to find x so that y = 7 = 2x + 1. That's easy, you just solve for x, so let's do it in general. If you give me a y and ask for x so that y = 2x + 1 I subtract 1 from each side and then divide by 2 to get x = (y - 1)/2. Hence no matter what real number y you give me I can find an x so that y = 2x + 1. Thus the range of y = 2x + 1 ia all real numbers.

Finally look at f(x) = (8x-3)/(4x-1). What can go wrong with this function? There is no square root to worry about but there is a fraction and I know that for fractions you can't have a denominator that is zero. Hence this function breaks if the denominator is zero, that is 4x - 1 = 0. Solving for x gives x = 1/4. This is the only value of x that breaks the function so the domain of f(x) = (8x-3)/(4x-1) is al real numbers x except x = 1/4. What about the range? Again the challenge, you give me a number y and ask for x so that y = (8x-3)/(4x-1). This time as before I need to solve for x to see if such an x exists. When I solved for x I got x = (y-3)/(4y-2). Again a fraction and I see a problem if the y you gave me was 1/2 because then the denominator is zero. Every other value of y works so the range of f(x) = (8x-3)/(4x-1) is all real numbers y except y = 1/2.

These three problems are somewhat misleading. For most functions y = f(x) if you give me a y I can't solve for x.

I hope this helps,
Penny