# Solving range algebraically

• May 7th 2013, 07:25 PM
symphonicity
Solving range algebraically
Let's say you have a function in which you have a linear function divided by a parabola. How do you find the range of this? Also, how do you find the range when you have something such as a parabola divided by a parabola? Thanks for your help.
• May 7th 2013, 10:26 PM
chiro
Re: Solving range algebraically
Hey symphonicity.

The answer will depend on the nature of the coefficients, but as a guide you should find out:

a) When the denominator becomes zero (if it becomes zero)
b) What the limit of the function is at plus or minus infinity
c) When the first derivatives are zero (and what kind of points they are)

These should give the answers to get the range.
• May 7th 2013, 11:50 PM
Prove It
Re: Solving range algebraically
Quote:

Originally Posted by symphonicity
Let's say you have a function in which you have a linear function divided by a parabola. How do you find the range of this? Also, how do you find the range when you have something such as a parabola divided by a parabola? Thanks for your help.

A nice method is to find the domain of the inverse relation, if it's possible to write the inverse relation as y in terms of x...
• May 8th 2013, 05:19 AM
HallsofIvy
Re: Solving range algebraically
The "range" of function y= f(x) is the set of all values of y given by some value of x in the domain. What you can do is look for the x that will give a value of y. And that is equivalent to finding the inverse function as Prove It suggested.

For example, to find the range of $f(x)= x^2- 3x+ 4$, we can write $y= x^2- 3x+ 4$ and attempt to solve for x. Since this particular function is quadratic, $x^2- 3x+ (4-y)= 0$, the quadratic formula works nicely: $x= \frac{3\pm\sqrt{9- 4(4- y)}}{2}= \frac{3\pm\sqrt{4y- 7}}{2}$. There will exist real x satisfying that as long as $4y- 7\ge 0$ so the range is "all y greater than or equal to 7/4".