Solving Equations and Inequalities Graphically

Some equations can’t be solved with algebra at all. Try isolating $x$ in $2^x = x + 3$: taking logs doesn’t help, because $x$ is stuck both in an exponent and outside one. Equations like this, which mix different function types, come up all the time in real problems. You can still solve them, very accurately, by using graphs to find roughly where the solutions are and then narrowing in with numbers. Trig functions here use radians.

Key ideas

Two ways to use a graph

To solve $f(x) = g(x)$:

1. Graph both sides. Graph $y = f(x)$ and $y = g(x)$ on the same axes. The solutions are the $x$-coordinates of the points where the graphs intersect.
2. Find zeros. Move everything to one side: $h(x) = f(x) - g(x) = 0$. Graph $y = h(x)$. The solutions are its zeros ($x$-intercepts).

Both give the same answers. The first is good for seeing how many solutions there are and roughly where. The second is handy for narrowing them down, because you only have to watch one sign.

Narrowing in with a sign change

If $h(a)$ and $h(b)$ have opposite signs (and $h$ has no breaks between them), the graph must cross the $x$-axis somewhere between $a$ and $b$. So you can trap a solution:

1. Find two values where $h$ changes sign, using the graph or a table.
2. Test a value in between, and keep the half where the sign still changes.
3. Repeat until the interval is small enough.

To round to 2 decimal places, show a sign change across the rounding interval. For example, if $h(2.435)$ and $h(2.445)$ have opposite signs, the solution is between them, so it rounds to $2.44$.

Inequalities

To solve $f(x) \lt g(x)$, find where the graphs intersect first. Those points split the $x$-axis into intervals, and on each interval one graph stays below the other. The solution is the set of intervals where the graph of $f$ is below the graph of $g$. (This is the same idea as polynomial inequalities, but the boundary points now come from a graph instead of from factoring.)

Technology

Graphing technology such as Desmos finds intersection points and zeros instantly, and it’s a great way to check your work. But you should always be able to explain where the answer came from: an intersection, a zero, or a sign change. A by-hand table with a calculator works anywhere.

Worked examples

Example 1: An exponential and a line

Solve $2^x = x + 3$. Give the answers to 2 decimal places.

Solution. Graph $y = 2^x$ and $y = x + 3$.

$y = 2^x$ and $y = x + 3$ cross twice, so the equation has two solutions.

The graphs cross twice: once between $x = -3$ and $x = -2$, and once between $x = 2$ and $x = 3$. A line can’t cross an exponential curve more than twice, so these are all the solutions.

Let $h(x) = 2^x - x - 3$ and narrow in on each crossing (values to 4 decimal places).

$x$	$-3$	$-2.9$	$-2.865$	$-2.86$	$-2.855$	$2.4$	$2.435$	$2.44$	$2.445$	$2.5$
$h(x)$	$0.125$	$0.0340$	$0.0023$	$-0.0023$	$-0.0068$	$-0.1220$	$-0.0274$	$-0.0136$	$0.0003$	$0.1569$

• $h(-2.865) \gt 0$ and $h(-2.855) \lt 0$, so the first solution is between them and rounds to $x \approx -2.86$.
• $h(2.435) \lt 0$ and $h(2.445) \gt 0$, so the second solution rounds to $x \approx 2.44$.

Check: $2^{2.44} \approx 5.43$ and $2.44 + 3 = 5.44$. They agree to within rounding. ✓

Example 2: Cosine equals x

Solve $\cos x = x$, where $x$ is in radians. Give the answer to 2 decimal places.

Solution. Sketch $y = \cos x$ and $y = x$. The line $y = x$ is above $1$ for $x \gt 1$, and below $-1$ for $x \lt -1$, so any crossing must have $-1 \le x \le 1$. For $-1 \le x \le 0$, $\cos x$ is positive while $x$ is not, so there’s no crossing there either. Between $0$ and $1$, $\cos x$ falls from $1$ while $y = x$ rises from $0$, so they cross exactly once.

Let $h(x) = \cos x - x$ and narrow in (calculator in radian mode):

$x$	$0.7$	$0.73$	$0.735$	$0.74$	$0.745$	$0.75$
$h(x)$	$0.0648$	$0.0152$	$0.0068$	$-0.0015$	$-0.0099$	$-0.0183$

The sign changes between $0.735$ and $0.745$ (in fact between $0.735$ and $0.74$), so the solution rounds to

$$x \approx 0.74$$

Check: $\cos 0.74 \approx 0.738$, very close to $0.74$. ✓

Example 3: An inequality

Solve $2x^2 \lt 2^x$. Give the boundary values to 2 decimal places.

Solution. First solve the equation $2x^2 = 2^x$. Graph both sides.

$2^x$ is above $2x^2$ between the first two crossings and after the third.

The graphs meet three times. One crossing is exact: at $x = 1$, both sides equal $2$. Narrow in on the other two with $h(x) = 2x^2 - 2^x$:

$x$	$-0.585$	$-0.575$	$6.315$	$6.325$
$h(x)$	$0.0178$	$-0.0100$	$0.1417$	$-0.1593$

So the crossings are at $x \approx -0.58$, $x = 1$ and $x \approx 6.32$.

Now test a value in each interval to see where $2x^2 \lt 2^x$:

Interval	Test $x$	$2x^2$	$2^x$	$2x^2 \lt 2^x$?
$x \lt -0.58$	$-1$	$2$	$0.5$	no
$-0.58 \lt x \lt 1$	$0$	$0$	$1$	yes
$1 \lt x \lt 6.32$	$2$	$8$	$4$	no
$x \gt 6.32$	$7$	$98$	$128$	yes

The solution is approximately

$$-0.58 \lt x \lt 1 \quad\text{or}\quad x \gt 6.32$$

Notice the last interval: it’s easy to miss if your graph stops at $x = 5$. Because exponentials eventually beat polynomials, $2^x$ must overtake $2x^2$ again somewhere.

Example 4: Two growing towns

Town A has $8000$ people and grows by $4\%$ a year, so after $t$ years its population is $A(t) = 8000(1.04)^t$. Town B has $12\,000$ people and grows by $100$ people a year: $B(t) = 12\,000 + 100t$. When will the towns have the same population? Answer to 2 decimal places.

Solution. Solve $8000(1.04)^t = 12\,000 + 100t$. This mixes an exponential and a linear function, so use a graph or a table. Let $h(t) = A(t) - B(t)$:

$t$	$10$	$12$	$12.945$	$12.955$	$13$
$h(t)$	$-1158.0$	$-391.7$	$-2.6$	$1.6$	$20.6$

$h$ changes sign between $12.945$ and $12.955$, so the populations are equal after about $t \approx 12.95$ years, when both towns have about $13\,300$ people. Before that Town B is bigger, and after it Town A’s exponential growth keeps it ahead for good.

Common mistakes

Trying to solve with algebra that can’t work. Taking logs of $2^x = x + 3$ gives $x\log 2 = \log(x + 3)$, which still has $x$ in two different kinds of places. Recognize these equations early and switch to a graph.

Stopping after the first solution. Graph over a wide enough window to see every crossing. In Example 3, the crossing near $6.32$ is far from the others.

Rounding from a single nearby value. Seeing that $h(2.44)$ is small doesn’t prove the answer rounds to $2.44$. Show a sign change between $2.435$ and $2.445$.

Using degree mode. In $\cos x = x$, $x$ is a real number of radians. In degree mode you’ll get a completely different (and wrong) answer.

Reading the wrong intervals for an inequality. $f(x) \lt g(x)$ is where the graph of $f$ is below the graph of $g$. Test a point in each interval to be sure.

Giving the y-coordinate as the solution. The solutions of $f(x) = g(x)$ are the $x$-coordinates of the intersection points. The $y$-coordinate is just the common value.

Practice

1. (Warm-up) The equation $3^x = 5 - x$ can’t be solved algebraically.

• (a) Explain why it has exactly one solution.
• (b) Show that the solution lies between $1$ and $2$, then find it to 2 decimal places.

Solution

(a) $y = 3^x$ is always increasing and $y = 5 - x$ is always decreasing, so their graphs can cross at most once. And they do cross, since $3^x$ starts below the line on the left and ends up above it on the right.

(b) Let $h(x) = 3^x + x - 5$. Then $h(1) = 3 + 1 - 5 = -1 \lt 0$ and $h(2) = 9 + 2 - 5 = 6 \gt 0$, so there’s a solution between $1$ and $2$.

Narrowing in: $h(1.2) \approx -0.0628$, $h(1.205) \approx -0.0372$, $h(1.215) \approx 0.0143$. The sign changes between $1.205$ and $1.215$, so $x \approx 1.21$.

2. (Warm-up) Show that $x^3 + x = 7$ has a solution between $1$ and $2$, and find it to 2 decimal places.

Solution

Let $h(x) = x^3 + x - 7$. Then $h(1) = -5$ and $h(2) = 3$: a sign change.

$x$	$1.7$	$1.8$	$1.73$	$1.735$	$1.745$
$h(x)$	$-0.387$	$0.632$	$-0.092$	$-0.042$	$0.059$

The sign changes between $1.735$ and $1.745$, so $x \approx 1.74$.

3. (Warm-up) Without solving, explain how many solutions $\sin x = 0.3x$ has ($x$ in radians). Use a sketch of both sides.

Solution

$\sin x$ is always between $-1$ and $1$, and $0.3x$ is between $-1$ and $1$ only for $-3.33 \le x \le 3.33$. So all solutions are in that interval.

$x = 0$ is a solution. For $x \gt 0$: $\sin x$ starts out rising faster than the line (at $x = 2$, $\sin 2 \approx 0.91 \gt 0.6$), but by $x = 3$ it’s below it ($\sin 3 \approx 0.14 \lt 0.9$), so they cross once between $2$ and $3$. For $\pi \lt x \le 3.33$, $\sin x$ is negative while $0.3x$ is positive, so there are no more crossings. Both sides are odd functions, so the picture for $x \lt 0$ is the mirror image.

There are $3$ solutions: $x = 0$ and one each near $\pm 2.4$. (Narrowing in gives $\pm 2.36$.)

4. (Core) Solve $x^3 - 3x + 1 = 0$. Give all solutions to 2 decimal places.

Solution

The rational root candidates $\pm 1$ don’t work ($1 - 3 + 1 = -1$ and $-1 + 3 + 1 = 3$), so it doesn’t factor nicely. Use a table of $h(x) = x^3 - 3x + 1$ to find sign changes:

$x$	$-2$	$-1$	$0$	$1$	$2$
$h(x)$	$-1$	$3$	$1$	$-1$	$3$

There are sign changes in $(-2, -1)$, $(0, 1)$ and $(1, 2)$: three solutions, the most a cubic can have. Narrowing in:

• $h(-1.885) \approx -0.043$ and $h(-1.875) \approx 0.033$, so $x \approx -1.88$.
• $h(0.345) \approx 0.006$ and $h(0.355) \approx -0.020$, so $x \approx 0.35$.
• $h(1.525) \approx -0.028$ and $h(1.535) \approx 0.012$, so $x \approx 1.53$.

5. (Core) Solve $3^x \gt x + 2$. Give boundary values to 2 decimal places where needed.

Solution

Solve $3^x = x + 2$ first. $x = 1$ is exact: $3 = 3$. A line meets an exponential at most twice; let $h(x) = 3^x - x - 2$. $h(-2) \approx 0.111 \gt 0$ and $h(0) = -1 \lt 0$, so there’s another crossing between $-2$ and $0$. Narrowing in: $h(-1.875) \approx 0.0025$ and $h(-1.865) \approx -0.0061$, so it’s at $x \approx -1.87$.

Test each interval:

• $x = -2$: $h \gt 0$, so $3^x \gt x + 2$. ✓
• $x = 0$: $h = -1 \lt 0$. ✗
• $x = 2$: $h = 9 - 4 = 5 \gt 0$. ✓

The solution is approximately $x \lt -1.87$ or $x \gt 1$.

6. (Core) Solve $\log x = 2 - x$. Give the answer to 2 decimal places.

Solution

$\log x$ is increasing and $2 - x$ is decreasing, so there’s at most one solution, and the domain needs $x \gt 0$. Let $h(x) = \log x + x - 2$.

$h(1) = 0 + 1 - 2 = -1$ and $h(2) = \log 2 \approx 0.301$: a sign change. Narrowing in:

$x$	$1.7$	$1.8$	$1.75$	$1.76$	$1.755$	$1.765$
$h(x)$	$-0.070$	$0.055$	$-0.007$	$0.006$	$-0.0007$	$0.012$

The sign changes between $1.755$ and $1.765$, so $x \approx 1.76$.

7. (Core) A town has $15\,000$ people and grows by $3\%$ a year. Its water system can serve $18\,000 + 150t$ people after $t$ years, as it’s gradually upgraded. After how many years will the population exceed the system’s capacity? Answer to 2 decimal places.

Solution

Solve $15\,000(1.03)^t = 18\,000 + 150t$. Let $h(t) = 15\,000(1.03)^t - 18\,000 - 150t$.

$t$	$5$	$10$	$8$	$9$
$h(t)$	$-1361$	$659$	$-198$	$222$

Narrowing in between $8$ and $9$: $h(8.47) \approx -3.1$ and $h(8.48) \approx 1.1$. Checking the rounding interval: $h(8.475) \approx -1.0$ and $h(8.485) \approx 3.2$, so $t \approx 8.48$.

The population exceeds the system’s capacity after about $8.48$ years, that is, during the ninth year.

8. (Challenge) Solve $\cos x \ge x^2$, with $x$ in radians. Give the boundary values to 2 decimal places.

Solution

Both sides are even functions, so the graph is symmetric about the $y$-axis. At $x = 0$, $\cos 0 = 1 \gt 0$. As $x$ moves away from $0$, $\cos x$ decreases and $x^2$ increases, and for $\lvert x \rvert \ge 1$, $x^2 \ge 1 \ge \cos x$ (with equality impossible, since $\cos 1 \lt 1$). So there’s exactly one crossing with $0 \lt x \lt 1$ and its mirror image.

Let $h(x) = \cos x - x^2$. $h(0.815) \approx 0.0216$ and $h(0.825) \approx -0.0021$, so the crossing is at $x \approx 0.82$, and by symmetry at $x \approx -0.82$.

Since $h(0) = 1 \gt 0$, the inequality holds between the crossings:

$$-0.82 \le x \le 0.82 \quad (\text{approximately})$$

9. (Challenge) Find all solutions of $x = 2\sin x$, with $x$ in radians, to 2 decimal places.

Solution

$2\sin x$ is between $-2$ and $2$, so any solution has $-2 \le x \le 2$. $x = 0$ is a solution. Both sides are odd, so solutions come in pairs $\pm x$.

For $x \gt 0$, let $h(x) = x - 2\sin x$. Near $0$, $2\sin x$ rises faster than $x$ (for example, $h(1) = 1 - 2\sin 1 \approx -0.68$), but $h(2) = 2 - 2\sin 2 \approx 0.18$. So there’s a crossing between $1$ and $2$.

$x$	$1.8$	$1.9$	$1.895$	$1.905$
$h(x)$	$-0.148$	$0.007$	$-0.0008$	$0.016$

The sign changes between $1.895$ and $1.905$, so $x \approx 1.90$.

The solutions are $x = 0$ and $x \approx \pm 1.90$.