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[Post New]14 Aug 2007 15:22:15 IST
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nivedh_89 (4653)

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Solving Polynomial Inequalities by Graphing

This section assumes that you have access to a graphing calculator or some other graphing program.
Let's suppose you want to solve the inequality
x2-1<0.

Here is the graph of the function f(x)=x2-1:
A given x will solve the inequality if f(x)<0, i.e., if f(x) is below the x-axis. Thus the set of our solutions is the part of the x-axis indicated below in red, the interval (-1,1):
If we want to see the solutions of the inequality
x2-1>0,

that's just as easy. Now we have to pick all values of x for which f(x)=x2-1 is above the x-axis. As you can see, we obtain as solutions the set $(-infty,1)cup(1,infty)$, indicated below in blue.
Note the pivotal role played by the "yellow dots", the x-intercepts of f(x).
f(x) can only change its sign by passing through an x-intercept, i.e., a solution of f(x)=0 will always separate parts of the graph of f(x) above the x-axis from parts below the x-axis. This property of polynomials is called the Intermediate Value Property of polynomials; your teacher might also refer to this property as continuity.
Let us consider another example: Solve the inequality
egin{displaymath}x^4+x^3-2x^2-2xgeq 0.end{displaymath}

Here is the graph of the function f(x)=x4+x3-2x2-2x>0:
A given x will solve the inequality if $f(x)geq 0$, i.e., if f(x) is above the x-axis. Thus the set of our solutions is the part of the x-axis indicated below in blue, the union of the following three intervals:
egin{displaymath}(-infty,-1.4]cup [-1,0] cup [1.4,infty)end{displaymath}

The (finite) endpoints are included since at these points f(x)=0 and so these x's are included in our quest of finding the solutions of $f(x)geq 0$.
Our answer is approximate, the endpoints of the intervals were found by inspection; you can usually obtain better estimates for the endpoints by using a numerical solver to find the solutions of f(x)=0. In fact, as you will learn in the next section, the precise endpoints of the intervals are $-sqrt{2}approx -1.414$, -1, 0 and $sqrt{2}approx 1.414$.
Two more caveats: The method will only work, if your graphing window contains all x-intercepts. Here is a rather simple-minded example to illustrate the point: Suppose you want to solve the inequality
x2-10x<0.

If your graphing window is set to the interval [-5,5], you will miss half of the action, and probably come up with the incorrect answer:
To find the correct answer, the interval (0,10), your graphing window has to include the second x-intercept at x=10:
Here is another danger: Consider the three inequalities $x^2 leq 0$, $x^2-0.0001leq 0$ and $x^2+0.0001leq 0$. If you do not zoom in rather drastically, all three graphs look about the same:
Only zooming in reveals that the solutions to the three inequalities show a rather different behavior. The first inequality has a single solution, x=0. (This also illustrates the fact that a function f(x) does not always change sign at points where f(x)=0.)
The second inequality, $x^2-0.0001leq 0$, has as its solutions the interval [-0.01,0.01]:
The third inequality, $x^2+0.0001leq 0$, has no solutions:


now.........Solving Polynomial Inequalities Analytically


Let's suppose you want to solve the inequality
x2-4x+3<0.

Step 1. Solve the equation f(x)=x2-4x+3=0.
In this case, we can "factor by guessing":
x2-4x+3=(x-1)(x-3),

so the roots of the equation f(x)=0 are x=1 and x=3. Draw a picture of the x-axis and mark these points.
Step 2. Our solutions partition the x-axis into three intervals. Pick a point (your choice!) in each interval. Let me take x=0,x=2 and x=4. Compute f(x) for these points:
egin{eqnarray*}f(0)&=&0-0+3=3>0\ f(2)&=&4-8+3<0\ f(4)&=&16-16+3>0 end{eqnarray*}

These three points are representative for what happens in the intervals they are contained in:
Since f(0)>0, f(x) will be positive for all x in the interval $(-infty,1)$. Similarly, since f(2)<0, f(x) will be negative for all x in the interval (1,3). Since f(4)>0, f(x) will be positive for all x in the interval $(3,infty)$. You can indicate this on the x-axis by inserting plus or minus signs on the x-axis. I use color coding instead: blue for positive, red for negative:
Step 3. We want to solve the inequality
x2-4x+3<0,

so we are looking for all x such that f(x)<0. Consequently, the interval (1,3) contains all solutions to the inequality.
Why does this work? Let's look at the graph of f(x):
Note the pivotal role played by the "yellow dots", the x-intercepts of f(x).
f(x) can only change its sign by passing through an x-intercept, i.e., a solution of f(x)=0 will always separate parts of the graph of f(x) above the x-axis from parts below the x-axis. Thus it suffices to pick a representative in each of the three intervals separated by "yellow dots", to test whether f(x) is positive or negative in the interval .
This nice property of polynomials is called the Intermediate Value Property of polynomials; your teacher might also refer to this property as continuity.
Here is another example: Find the solutions of the inequality
egin{displaymath}x^2+2geq 3x.end{displaymath}

For our method to work it is essential that the right side of the inequality equals zero! So let's change our inequality to
egin{displaymath}x^2-3x+2geq 0.end{displaymath}

Step 1. Solve the equation f(x)=x2-3x+2=0.
Again, we can "factor by guessing":
x2-3x+2=(x-1)(x-2),

so the roots of the equation f(x)=0 are x=1 and x=2. Draw a picture of the x-axis and mark these points.
Step 2. Our solutions partition the x-axis into three intervals. Pick a point (your choice!) in each interval. Let me take x=0,x=1.5 and x=3. Compute f(x) for these points:
egin{eqnarray*}f(0)&=&0-0+2>0\ f(1.5)&=&2.25-4.5+2<0\ f(3)&=&9-9+2>0 end{eqnarray*}

These three points are representative for what happens in the intervals they are contained in:
Since f(0)>0, f(x) will be positive for all x in the interval $(-infty,1)$. Similarly, since f(1.5)<0, f(x) will be negative for all x in the interval (1,2). Since f(3)>0, f(x) will be positive for all x in the interval $(2,infty)$. You can indicate this on the x-axis by inserting plus or minus signs on the x-axis. I use color coding instead: blue for positive, red for negative:
Step 3. We want to solve the inequality
egin{displaymath}x^2-3x+2geq 0,end{displaymath}

so we are looking for all x such that $f(x)geq 0$. Consequently, the set $(-infty,1]cup [2,infty)$ contains all solutions to the inequality. (Since our inequality only stipulates that $f(x)geq 0$, x=1 and x=2 are solutions, so we include them. "$+infty$" and "$-infty$" are only symbols; they will never be included as solutions.)
Our next example: Solve x3>2x. Do not divide by x on both sides! If you do so, you will never be able to arrive at the correct answer. Repeat the pattern instead; make one side of the inequality equal zero:
x3-2x>0.

Step 1. Solve the equation f(x)=x3-2x=0.
We can factor rather easily:
egin{displaymath}x^3-2x=x(x^2-2)=x(x-sqrt{2})(x+sqrt{2}),end{displaymath}

so the roots of the equation f(x)=0 are $x=-sqrt{2}$, x=0 and $x=sqrt{2}$. Draw a picture of the x-axis and mark these points.
Step 2. Our solutions partition the x-axis into four intervals. Pick a point (your choice!) in each interval. Let me take x=-2,x=-1, x=1 and x=2. Compute f(x) for these points:
egin{eqnarray*}f(-2)&=&-8+4<0\ f(-1)&=&-1+2>0\ f(1)&=&1-2<0\ f(2)&=&8-4>0 end{eqnarray*}

These four points are representative for what happens in the intervals they are contained in:
Since f(-2)<0, f(x) will be negative for all x in the interval $(-infty,-sqrt{2})$. Similarly, since f(-1)>0, f(x) will be positive for all x in the interval $(-sqrt{2},0)$. Since f(1)<0, f(x) will be negative for all x in the interval $(0,sqrt{2})$. Since f(2)>0, f(x) will be positive for all x in the interval $(sqrt{2},infty)$. You can indicate this on the x-axis by inserting plus or minus signs on the x-axis. I use color coding instead: blue for positive, red for negative:
Step 3. We want to solve the inequality
x3-2x> 0,

so we are looking for all x such that f(x)> 0. Consequently, the set $(-sqrt{2},0)cup (sqrt{2},infty)$ contains all solutions to the inequality. (Since our inequality stipulates that f(x)>0, $x=pmsqrt{2}$ do not qualify as solutions, so we exclude them. "$+infty$" and "$-infty$" are only symbols; they will never be included as solutions.)
Here is my last example: Solve
egin{displaymath}x^3+3x^2+x+3leq 0.end{displaymath}

Step 1. Solve the equation f(x)=x3+3x2+x+3.
We can factor either by finding a rational zero, or by clever grouping:
x3+3x2+x+3=(x3+3x2)+(x+3)=x2(x+3)+1 (x+3)=(x2+1)(x+3),

so there is only one real root of the equation f(x)=0, namely x=-3. Draw a picture of the x-axis and mark this point.
The polynomial x2+1 is irreducible, it does not have real roots. Its complex roots are irrelevant for our purposes.
Step 2. Our solution partitions the x-axis into two intervals. Pick a point (your choice!) in each interval. Let me take x=-4 and x=0. Compute f(x) for these points:
egin{eqnarray*}f(-4)&=&-17<0\ f(0)&=&3>0 end{eqnarray*}

These two points are representative for what happens in the intervals they are contained in:
Since f(-4)<0, f(x) will be negative for all x in the interval $(-infty,-3)$. Similarly, since f(0)>0, f(x) will be positive for all x in the interval $(-3,infty)$. You can indicate this on the x-axis by inserting plus or minus signs on the x-axis. I use color coding instead: blue for positive, red for negative:
Step 3. We want to solve the inequality
egin{displaymath}x^3+3x^2+x+3leq 0,end{displaymath}

so we are looking for all x such that $f(x)leq 0$. Consequently, the set $(-infty,-3]$ is the set of solutions to the inequality. (Since our inequality stipulates that $f(x)leq 0$, x=-3 is also a solution, so we include it. "$+infty$" and "$-infty$" are only symbols; they will never be included as solutions.)
The inevitable truth of life.....everyone in our life is going 2 hurt sooner or later......u just have 2 realise who is worth.....

the PAIN or the PERSON...!!!
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