Solving Inequalities
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Solve each of the following inequalities.
1.
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To solve a polynomial inequality
we get a zero on one side of the inequality, factor and then determine where
the other side is zero.
So, once we move everything over
to the left side and factor we can see that the left side will be zero at and . These numbers are NOT solutions (since we
only looking for values that will make the equation positive) but are useful to
finding the actual solution.
To find the solution to this
inequality we need to recall that polynomials are nice smooth functions that
have no breaks in them. This means that
as we are moving across the number line (in any direction) if the value of the
polynomial changes sign (say from positive to negative) then it MUST go through
zero!
So, that means that these two
numbers ( and ) are the ONLY places where the polynomial
can change sign. The number line is then
divided into three regions. In each
region if the inequality is satisfied by one point from that region then it is
satisfied for ALL points in that region.
If this wasn’t true (i.e it
was positive at one point in region and negative at another) then it must also
be zero somewhere in that region, but that can’t happen as we’ve already
determined all the places where the polynomial can be zero! Likewise, if the inequality isn’t satisfied
for some point in that region that it isn’t satisfied for ANY point in that
region.
This means that all we need to do
is pick a test point from each region (that are easy to work with, i.e. small integers if possible) and
plug it into the inequality. If the test
point satisfies the inequality then every point in that region does and if the
test point doesn’t satisfy the inequality then no point in that region
does.
One final note here about
this. I’ve got three versions of the
inequality above. You can plug the test
point into any of them, but it’s usually easiest to plug the test points into
the factored form of the inequality. So,
if you trust your factoring capabilities that’s the one to use. However, if you HAVE made a mistake in
factoring, then you may end up with the incorrect solution if you use the
factored form for testing. It’s a
trade-off. The factored form is, in many
cases, easier to work with, but if you’ve made a mistake in factoring you may
get the incorrect solution.
So, here’s the number line and
tests that I used for this problem.
From this we see that the solution
to this inequality is and . In interval notation this would be and . You’ll notice that the endpoints were not
included in the solution for this. Pay
attention to the original inequality when writing down the answer for these. Since the inequality was a strict inequality,
we don’t include the endpoints since these are the points that make both sides
of the inequality equal!
2.
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3.
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4.
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The process for solving
inequalities that involve rational functions is nearly identical to solving
inequalities that involve polynomials.
Just like polynomial inequalities, rational inequalities can change sign
where the rational expression is zero.
However, they can also change sign at any point that produces a division
by zero error in the rational expression.
A good example of this is the rational expression . Clearly, there is division by zero at and to the right of the expression is positive and to the left
of the expression is negative.
It’s also important to note that a
rational expression will only be zero for values of x that make the numerator zero.
So, what we need to do is first
get a zero on one side of the inequality so we can use the above
information. For this problem that has
already been done. Now, determine where
the numerator is zero (since the whole expression will be zero there) and where
the denominator is zero (since we will get division by zero there).
At this point the process is
identical to polynomial inequalities with one exception when we go to write
down the answer. The points found above
will divide the number line into regions in which the inequality will either
always be true or always be false. So,
pick test points from each region, test them in the inequality and get the
solution from the results.
For this problem the numerator
will be zero at and the denominator will be zero at . The number line, along with the tests is
shown below.
So, from this number line it looks
like the two outer regions will satisfy the inequality. We need to be careful with the endpoints
however. We will include because this will make the rational expression
zero and so will be part of the solution.
On the other hand, will give division by zero and so MUST be
excluded from the solution since division by zero is never allowed.
The solution to this inequality is
and OR and ,
depending on if you want inequality for the solution or intervals for the
solution.
5.
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6.
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