Find all open intervals on which f(x)=x/x^2+x-2 is decreasing.

Question

What are the answers? I know the answers ,but I don`t know how to attempt the problem to get them. What is the difference between a closed and an open interval for this problem? Thanks.

Answer 1

f(x) = x/(x²+x-2)

 

Domain = {x l x²+x-2≠0} = {x l (x+2)(x-1)≠0} = {x l x ≠ -2, 1}

 

f(x) is decreasing for all values of x where f'(x) < 0

 

By the quotient rule, f'(x) = [(x²+x-2)(1)-x(2x+1)]/(x²+x-2)²

 

                                    = (-x²-2)/(x²+x-2)²

 

                                    = -(x²+2)/(x²+x-2)² < 0 for all x ≠ -2, 1

 

f(x) is decreasing on (-∞, -2)∪(-2,1)∪(1,∞)

 

 

Answer 2

 

Hi Ab,

 

Note that the derivative of a quotient, g(x) = p(x)/q(x), is

 

     g'(x) = (p'(x)q(x) - p(x)q'(x))/(q(x))².

 

Applying this rule to f(x), we get

 

     f'(x) = ((x)'(x²+x-2) - x(x²+x-2)')/(x²+x-2)²

 

            = ((x²+x-2) - x(2x+1))/(x²+x-2)²

 

            = (-x²-2)/(x²+x-2)²

 

            = -(x²+2)/(x²+x-2)²

 

As f'(x) is negative,

 

f(x) is decreasing except when x = -2 or x = 1,

 

So, the intervals on which f(x) is decreasing are

 

     (-∞, -2), (-2, 1), (1, ∞)

 

 

 

Answer 3

Hi this is the problem you posted f(x)=x/x^2+x-2

There is one small concern with the way you posted, this problem can be interpreted as either x/x^2... x -2 

or x/(x^2+x-2) and there is a huge difference because the denominator is what determines the domain for that function.

We want to first find out the values of x when the denominator/s is/are = 0.

I'm going to assume that your problem is x/(x^2+x-2) in which case we have to factor the denominator part.

x^2+x-2 factors to (x+2)(x-1) and when these are set to 0, x = -2,1.

These two x values shows what would turn the denominator into 0 if you were to plug them back into the function.

Therefore your domain(values that are acceptable for your function) is going to be any numbers other than those two numbers. So the format you can enter for the domain is all real number when x is not equal to -2,1. Your question is asking for the open intervals when the function is decreasing, we are going to put open intervals on all values other than -2,1 such as (-infinity,-2),(-2,1),(1,+infinity) (answer) and the difference between closed and open intervals is that open (parenthesis) doesn't accepts numbers that are set as boundaries as domain (which is the reason why we use parenthesis since -2,1 are not acceptable as domains) whereas the closed ([]) accepts boundary numbers as domains as well. 

If you were to see the graphs of these particular types of functions (for open intervals) they are straight vertical lines, which means they are not acceptable as domains, or they may be shown as open dots. Whereas the ([]) may be shown on graph as closed dots or just a regular smooth graph with no holes.