Solve using the sign test. Express answers in interval notation.

Question

X+3/x-5.       -1 greater than or equal to0

Answer 1

The word SOLVE should be reserved for situations in which you are SOLVING AN EQUATION.

 

Here you have a rational algebraic expression, probably intended to have numerator (x+3) & denominator (x-5).

 

if we consider this to be f(x), the domain of x is all Reals except 5.

 

If we find all zeros, and place them on a numberline, we have

 

 

-------------------(-3)---------------------(5)-----------------------; the two zeros of the parts of this function have subdivided the numberline into three intervals.

 

The question then is, what is the sign of the function, in each interval?

 

Choose any number > 5; mentally f(6) = positive because both numerator & denominator are always positive.

 

Because both of the binomials within this problem are linear, the sign of the function will change when we go from the rightmost interval (5,infinity) into the adjacent interval (-3,5). 

And when moving further to the left to get into the interval (-infinity,-3) the sign of every function value will change again.T o summarize, the behavior of f is recorded beneath the intervals as:

 

------------------(-3)---------------------(5)-----------------------

f(x)>0              =0           f(x)<0        v.a.           f(x)>0

 

NOTE that isn't not necessary to actually compute values in any interval except the one.

 

If A FACTOR IS SQUARED (or has any even exponent) in a rational function, then at that point on the numberline there is NOT A SIGN CHANGE when moving from intervals containing said endpoont.

 

Answer 2

I'm not sure what your question is. The sign test is used to graph functions and solve inequalities. The function can be realized as a rational polynomial (x^2 + 5*x - 3)/x = (x-4.3028)*(x-0.9672)/x. To check as x->0 let x = .01, x=-.01

For x = .01 the signs are (-)(-)/(+) and for x = -.01, (-)(-)/(-). The function approaches plus inf from the right 0+ and negative infinity 0- from the left. A table of the function from x to f gives x = -2,-1,0-,0+,.5,1,2,3,4,5,6,20 -> f =  -8.5,-9,-inf, +inf, 1.5, -1, -1.5, -1, -.25, 0.6, 1.5, 15.15. From this you can sketch the function. This may help you answer your question.