Higher degree inequalities

Functions: Higher degree polynomials

Higher degree inequalities

In the same manner as solving a quadratic inequality, we can also solve an inequality with higher-degree polynomials.

Solving a higher-degree inequality

	Procedure	Example
	We solve the following inequality \[\blue{f(x)} \gt \green{g(x)}\] in which #\blue{f(x)}# and #\green{g(x)}# are polynomials.	#\blue{x^6+x^3+6} \gt \green{-2x^3+10}# (resp. solid and dashed) The solution is #x \lt \sqrt[3]{-4} \land x \gt 1#.
Step 1	We solve the equality \[\blue{f(x)} = \green{g(x)}\]
Step 2	We sketch the graphs #\blue{f(x)}# and #\green{g(x)}#.
Step 3	With the help of steps 1 and 2, we determine for which values of #x# the inequality holds. In a coordinate system, the biggest graph is the one above the other.

Please note that this procedure also holds for the inequality signs #\geq# and #\leq#, only now are the #x#-values of the intersection points also part of the solution.

All or noneIf no intersection points are found in step 1, there are two options for solving the inequality:

All points of #x# are a solution to the inequality. The inequality is always true.
No points of #x# are a solution to the inequality. The inequality is never true.

The same can happen if an intersection point is found in step 1, but when it's only a point of tangency. Then they are only equal in that point, and the inequality is true or not.

Alternative step 2

Instead of sketching the graphs, we can also do step #2# by substituting a few #x#-values in #\blue{f(x)}# and #\green{g(x)}#. This allows us to determine on which intervals the inequality #\blue{f(x)}\gt \green{g(x)}# holds.

A change in which function is greater can only occur at the intersection points of #\blue{f(x)}# and #\green{g(x)}#. Therefore, testing just one point in each interval between these intersection points is sufficient.

In the example above and on the right, the intersection points of #\blue{f(x)}# and #\green{g(x)}# are the #x# values #x = \sqrt[3]{-4}# and #x=1#. So, we choose one #x#-value smaller than # \sqrt[3]{-4}#, one #x#-value between #x= \sqrt[3]{-4}# and #x=1#, and one #x#-value larger than #1#. This gives us all the information we need to know when #\blue{f(x)}# is larger than #\green{g(x)}#.

Example

In the example above, we substitute a few #x#-values.

#\blue f(-2)=62 \gt \green g(-2)=26 #

#\blue f(0)=6 \lt\green g(0)=10#

#\blue f(2)=78 \gt \green g(2)=-6#

Step 3 then gives us that the solution is #x \lt \sqrt[3]{-4} \land x \gt 1#, because the inequality holds for #x=-2# and #x=2#.

Solve the inequality for #b#
\[b^6+4\cdot b^3+9\cdot b+43 \gt 9\cdot b+48\tiny\]
Give your answer in the form

#b \gt b_1\land b \lt b_2# or #b\lt b_1 \lor b \gt b_2#,
#b \lt b_1# or #b \gt b_1#,
#none# or #all#,

with the correct values of #b_1# and #b_2#.

#b\lt -5^{{{1}\over{3}}}\lor b\gt 1#

Step 1	We solve the equality #b^6+4\cdot b^3+9\cdot b+43=9\cdot b+48#. This is done like this: \[\begin{array}{rcl} b^6+4\cdot b^3+9\cdot b+43&=&9\cdot b+48 \\ &&\phantom{xxx}\blue{\text{original equation}}\\ b^6+4\cdot b^3-5&=&0 \\&&\phantom{xxx}\blue{\text{reduced to }0}\\ \left(b^3-1\right)\cdot \left(b^3+5\right)&=&0 \\&&\phantom{xxx}\blue{\text{factorized left-hand side}}\\ b^3-1=0 &\lor& b^3+5=0 \\&&\phantom{xxx}\blue{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\ b=-5^{{{1}\over{3}}} &\lor& b=1 \\&&\phantom{xxx}\blue{\text{constant terms to the right-hand side and taken the root}}\\ \end{array} \]
Step 2	We sketch the graphs #y=b^6+4\cdot b^3+9\cdot b+43# (blue) and #y=9\cdot b+48# (green dashed).
Step 3	We can read the solutions to the inequality from the graph. \[b\lt -5^{{{1}\over{3}}}\lor b\gt 1\]

New example