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Lesson 7 of 7Rates of Change & Combining Functions
Unit 6 · Lesson 7 of 7

Solving Equations and Inequalities Graphically and Numerically

When no algebra isolates , rewrite as , read the intersections with graphing technology, and pin each root down by sign changes and bisection to a stated accuracy — then read inequalities straight off the graph.

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Key things to remember

  1. 1Solve by finding the zeros of ; the zeros of are the -coordinates of the intersections.
  2. 2 exactly where , so the sign of between its zeros solves inequalities.
  3. 3If is continuous on and , have opposite signs, there is at least one root in .
  4. 4Bisection: test the midpoint, keep the half with the sign change; after steps the width is .
  5. 5Stop when the stated accuracy is guaranteed, for example when both endpoints round to the same value.
  6. 6Choose a graphing window from end behaviour so no solution is hidden off the screen, and use radians for trig equations.

Try to solve with algebra. Taking logarithms gives , which still has in two incompatible places. Try : one side is periodic and one is a straight line, and no identity connects them. These are not trick questions. Equations that mix families — an exponential with a polynomial, a trig function with a line, a logarithm with anything — usually have no algebraic solution at all, and they come up constantly when two models are compared: when does one town's population overtake another's, when does a cooling drink reach room temperature plus five degrees? Mathematicians do not give up on these. They switch to methods that find the answer to any accuracy you ask for, and those methods are the subject of this lesson.

Turning an equation into a zero-finding problem

Every equation can be rewritten with everything on one side. Define . Then exactly when , so the solutions of the equation are the zeros of — the -intercepts of a single graph. This one move pays off three ways. You only have to look at one curve and find where it meets the -axis, which is easier to read than where two curves cross. The sign of tells you which side is bigger: means , which is exactly what you need for inequalities. And a change of sign in between two inputs is a numerical signal that a solution lies between them, which is the basis of the bisection method.

Reading intersections with graphing technology

With a graphing calculator or graphing software you can either graph and and use the intersect feature, or graph and use the zero feature. The danger is the window. Graph and for and you will see the crossings at and and miss the third one entirely, near , because it is off the screen. Before trusting a graph, think about end behaviour: as , while grows, so is negative far to the left; at it is positive. A sign change between them guarantees another root. Check exact-looking answers by substitution — confirms — and report the others to the accuracy asked for, in radians for any trig equation.

x−2−1012345
−3.75−0.5110−107
Sign−−++0−0+
Sign changes between −1 and 0, at 2, and at 4: three solutions, and the signs give the inequality as well.

Sign changes and the bisection method

If is continuous on — its graph has no breaks — and and have opposite signs, the graph must cross the axis somewhere between them. Bisection turns that fact into an algorithm. Evaluate at the midpoint . If has the same sign as , the root is in ; otherwise it is in . Keep the half that still has a sign change and repeat. Every step halves the interval, so after steps an interval of width has shrunk to : ten halvings take a width of 1 below . You stop when the required accuracy is guaranteed — for example, when both endpoints round to the same value at the stated number of decimal places. Bisection is slow compared with a calculator's built-in solver, but it never fails on a continuous function with a sign change, and it is how you can be sure of every digit you report.

Bisection: solve cos x = x (x in radians) by halving a sign-change interval1. cos x = x: the graphs meet, but no algebra gives x00.511.50.51y = cos xy = xx = ?subtract x2. Rewrite as h(x) = cos x − x = 00.511−0.5rooth > 0h < 0h(0) = 1 > 0 and h(1) ≈ −0.460 < 0,so h changes sign: a root lies in [0, 1].3. Test the midpoint m; keep the half whose ends still have opposite signs00.250.50.751abmh(m)keep1+−010.5+0.378[m, b]2+−0.510.75−0.018[a, m]3+−0.50.750.625+0.186[m, b]4+−0.6250.750.6875+0.085[m, b]5+−0.68750.750.71875+0.034[m, b]After 5 halvings the root is in [0.71875, 0.75], an interval 1/32 wide. The root is x ≈ 0.7391.Each step halves the interval, so 10 halvings pin the root to within 1/1024, better than 0.001.
Fig 1Solving cos x = x as h(x) = cos x − x = 0: each midpoint test keeps the half of the interval where h still changes sign.
StepInterval [a, b]Midpoint mh(m) = cos m − mKeep
1[0, 1]0.50.3776 (+)[0.5, 1]
2[0.5, 1]0.75−0.0183 (−)[0.5, 0.75]
3[0.5, 0.75]0.6250.1860 (+)[0.625, 0.75]
4[0.625, 0.75]0.68750.0853 (+)[0.6875, 0.75]
5[0.6875, 0.75]0.718750.0339 (+)[0.71875, 0.75]
h(0) = 1 is positive and h(1) ≈ −0.4597 is negative. After five halvings the root lies in [0.71875, 0.75]; technology gives 0.7391.
VideoA university calculus video on the bisection method: keep halving an interval where the function changes sign until the root is pinned down.Watch on YouTube

Inequalities and applied problems

To solve , find the solutions of first; they are the only places where the inequality can switch from true to false. Then read the sign of on each interval between them, from the graph or by testing one point. For , the sign row above gives or (approximately). Endpoints are included for and excluded for . The same reasoning answers applied questions. Suppose Town A has 12 000 people growing by 3 percent per year, , and Town B has 18 000 people growing by 200 per year, . The equation mixes an exponential with a line, so you graph : it is negative at and positive at , and technology places the zero at years. So Town A is larger for — and because exponential growth eventually beats linear growth, it stays larger.

A sign change needs an unbroken graph
is negative at and positive at , yet it has no zero: the sign changes across the vertical asymptote at . Tangent does the same at its asymptotes. Bisection only finds roots of continuous functions, so check the interval for asymptotes or gaps before applying it, and confirm any root by substituting it back into the original equation.
VideoA Grade 12 lesson on techniques for solving equations and inequalities that algebra alone cannot handle.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1
Solve using graphing technology, then solve .
Example 2
Use five steps of bisection to locate the solution of (radians), starting on .
Example 3
Solve (base 10) to one decimal place for the solution between and , using bisection, and find how many solutions there are in total.
Example 4
Town A has people and Town B has , years from now. When will Town A become larger?

Where marks get lost

  • Using a window that starts at and missing a negative solution, such as for .
  • Keeping the wrong half in bisection: keep the half whose endpoints have OPPOSITE signs of .
  • Applying bisection across an asymptote, where the sign changes but there is no root.
  • Reporting a bisection midpoint to more decimal places than the final interval guarantees.
  • Solving a trig equation such as cos x = x with the calculator in degree mode.

Flashcards 9

Rewrite for a single graph
in terms of
What guarantees a root in [a, b]?
One step of bisection
+ 5 more

Practice

1
Which function's zeros are the solutions of ?
2
How many real solutions does have?
3
Find the negative solution of to 3 decimal places.
4
Bisection is applied to on . What midpoint is tested in the THIRD step?
5
Starting from an interval of width 1, how many bisection steps are needed to make the width less than 0.001?
steps
6
What is the name of the numerical method that repeatedly halves an interval containing a sign change?
7
Put one step of the bisection method for on in order.
  1. 1Check whether the interval is narrow enough for the required accuracy
  2. 2Keep the half-interval whose endpoints have opposite signs
  3. 3Evaluate h(m)
  4. 4Confirm h(a) and h(b) have opposite signs
  5. 5Compute the midpoint m = (a + b)/2
8
Match each equation to the function whose zeros solve it.
  1. A.
  2. B.
  3. C.
  4. D.
9
has zeros at about , and , and . Where is ?
10
Town A has people and Town B has , years from now. After how many years are they equal? Give 1 decimal place.
years
11
has and . What can you conclude?

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