Using Graph To Solve Inequality Of Rational Expressions

Question is here:

http://i39.tinypic.com/aexmjr.jpg

I didn't know how else to get the graph here. I don't even know where to start. My last unit was inequalities but I didn't 100% grasp the concept. What am I doing? Why is it necessary to use the graph?

Re: Using Graph To Solve Inequality Of Rational Expressions

You are given the graph of the function and then asked to solve the inequality when is . In other words, you are asked to find all values of where is a point on or below the -axis. So, where is that true? Let's look at the possibilities.

Looking at the graph makes it easy to rule out any answers that contradict the graph. For example, (b) is obviously wrong because -1 is a real number that is not equal to -2 nor 2, but is clearly above the -axis.

Among (a), (c), and (d), obviously one answer is correct and the rest are not. So, for the two that are not correct, tell me a point in the given intervals where the function is above the x-axis, or tell me a point that is not in the given intervals where the function is on or below the x-axis.

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Re: Using Graph To Solve Inequality Of Rational Expressions

Quote:

Originally Posted by

**tdotodot** Question is here:

http://i39.tinypic.com/aexmjr.jpg
I didn't know how else to get the graph here. I don't even know where to start. My last unit was inequalities but I didn't 100% grasp the concept. What am I doing? Why is it necessary to use the graph?

I highlighted the x-values for which f(x) is negative. which answer you will pic

Attachment 29485

Re: Using Graph To Solve Inequality Of Rational Expressions

Quote:

Originally Posted by

**votan** I highlighted the x-values for which f(x) is negative. which answer you will pic

Attachment 29485

So for that graph f(x) is negative when *-3<x<-2* and also *0>x<2*. Would this be the correct way to state it?

*also thanks for doing that graph.

Re: Using Graph To Solve Inequality Of Rational Expressions

Quote:

Originally Posted by

**tdotodot** So for that graph f(x) is negative when *-3<x<-2* and also *0>x<2*. Would this be the correct way to state it?

*also thanks for doing that graph.

You are not just asked where the graph f(x) is negative. You are also asked where it is zero. You are correct that it is negative on the interval -3 < x < -2. But, 0>x<2 means x<0 and x<2. I think the notation you were looking for is 0<x<2. Again, you are correct that f(x) is negative on that interval. Now, where is it zero?

Re: Using Graph To Solve Inequality Of Rational Expressions

The basic concept here is the "intermediate value property" of continuous functions: if f(a)< 0 and f(b)> 0 (or vice-versa) there exist c, between a and b, such that f(c)= 0. That says that f(x) can change from negative to positive or positive to negative only where f is 0 or where it is not continuous. For rational expressions, that can only occur where the numerator is 0 (f(c)= 0) or the denominator is 0 (f is not continuous). So the simplest way to solve, say, f(x)> 0, is to find all values of x where either the numerator or denominator is 0. Write those values in increasing order so (together with and ) they define a number of intervals then check f(x) at one value of x in each interval. If f(x)> 0 for that x, f is positive for all points in that interval and if f(x)< 0 for that x, f is negative for all points in that interval.

Re: Using Graph To Solve Inequality Of Rational Expressions

Quote:

Originally Posted by

**votan** I highlighted the x-values for which f(x) is negative. which answer you will pic

Attachment 29485

The convention in graphing is if a point is part of the interval it is marked with a black circle, a blank circle means that point is not part of the interva. You notice on the figure, the interval -3 to -2, point -3 is marked with a black circle, point -2 is blank circle. that means the interval is one sided open. One way to write it is [-3, -2), it is open on -2, closed on -3. another way of writing it is -3 <= x <-2. Does this help you answering SlipEternal question in the previous post?