Step By Step Calculus » 4.4 - Equalities and Inequalities

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Given two functions f(x)f(x) and g(x)g(x), we are interested in finding the exact solution set for an equation of the form f(x)=g(x)f(x)=g(x) and for inequalities of the form f(x)<g(x), f(x)\le g(x), f(x)>g(x), f(x)\ge g(x)f(x)<g(x), f(x)\le g(x), f(x)>g(x), f(x)\ge g(x).
Solving Equations: There are many methods for solving equations. The following are the most basic ones.
  • Isolating the Variable: To isolate a variable in an equation, we obey the following three-headed golden rule:
    • Seek simplicity above all
    • Do to the right side whatever you do to the left side
    • Track steps that may cause loss or gain of solutions. Make necessary adjustments or checks.
  • By Factoring: One can use the following procedure to solve many equations.
    Step 1: Move all terms to the left side, so that the right side is 00.
    Step 2: Factor the left side as much as possible.
    Step 3: Split the resulting equation into several smaller ones by setting each factor equal to 00.
    Step 4: Solve all the easier equations obtained in this way.
    Step 5: Check that all solutions are acceptable.
  • Involving Absolute Values: To solve equations involving absolute values, one can either consider all possible cases separately or apply squaring to the both sides. We can also use graphs, for example if we graph the left and right sides of an equation, then the solutions are the points of intersection of the graphs.
  • Using The Quadratic Formula: For a quadratic equation of the form ax^2+bx+c=0ax^2+bx+c=0, the solutions are \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. The solutions are real when the discriminant b^2-4ac\ge 0b^2-4ac\ge 0.
Solving Inequalities: The algebraic method of solving inequalities uses the order properties of the real numbers: If aa and bb are real numbers,
  • a<ba<b implies \displaystyle a+k<b+k\displaystyle a+k<b+k and a-k<b-ka-k<b-k for any k \in \mathbb{R}k \in \mathbb{R}.
  • a<ba<b and \displaystyle k>0\displaystyle k>0 implies ak < bk \; \; \forall k\in \mathbb{R}ak < bk \; \; \forall k\in \mathbb{R}.
  • a<ba<b and \displaystyle k<0\displaystyle k<0 implies ak>bk \; \; \forall k \in \mathbb{R}ak>bk \; \; \forall k \in \mathbb{R}.
  • a>0a>0 implies that \displaystyle \dfrac{1}{a}>0\displaystyle \dfrac{1}{a}>0
  • 0<a<b0<a<b implies that \displaystyle \dfrac{1}{a}>\dfrac{1}{b}\displaystyle \dfrac{1}{a}>\dfrac{1}{b}
  • 0 \leq a<b0 \leq a<b if and only if \displaystyle 0\leq a^{n}<b^{n}\displaystyle 0\leq a^{n}<b^{n} for any n\in\mathbb{N}n\in\mathbb{N}.
As a general strategy for solving inequalities of the form f(x)>g(x)f(x)>g(x) algebraically, we use the following COPS procedure.
Step 1: (Cut points) Find the values of xx for which either f(x)f(x) or g(x)g(x) are undefined or discontinuous and those for which f(x)=g(x)f(x)=g(x). We call these the cut points of inequality.
Step 2: (Order) Place these cut points on a number line in increasing order.
Step 3: (Pick and check) Pick a test value for xx in each of the intervals before, after or between the cut points and check whether for that value f(x)>g(x)f(x)>g(x) or not. The same conclusion will be true for any other value of xx in that same interval.
Step 4: (Solution) Choose the intervals for which the given inequality is satisfied.
If the inequality is of the form f(x)<g(x)f(x)<g(x) we can do exactly the same thing, except that in the third step the criterion for acceptability is the opposite.
Sometimes the inequality is of the form f(x)\ge g(x)f(x)\ge g(x) or f(x)\le g(x)f(x)\le g(x). In that case each of the cut points must be treated separately to see if it satisfies the inequality or not.
To solve inequalities, a graphical method may also be used, by plotting both functions carefully and visually determining where the equality or inequality holds. A disadvantage of this method is that it can be imprecise.
Triangle Inequality: The order properties of the real numbers can be used to prove the triangle inequality, which states that: \vert a+b\vert \leq \vert a\vert + \vert b \vert\vert a+b\vert \leq \vert a\vert + \vert b \vert for a,b\in\mathbb{R}a,b\in\mathbb{R}.