Continuity, question 21

Topic: Continuity
Difficulty: Standard
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §2.3 CLP-1 §1.6Key results

Show that the equation $\cos x = x$ has a solution in the interval $\left[0, \dfrac{\pi}{2}\right]$ .

Write your reasoning in full, then compare it against the model solution.

Your answer

Theory UBC Math 100 §2.3 CLP-1 §1.6Key results

Worked solution

Strategy

Turn the equation into a zero of one function, check that the function is continuous on the closed interval, and show that it changes sign between the ends. The Intermediate Value Theorem then gives a solution without finding it.

Working

$g(x) = \cos x - x$
cos x = x exactly where g(x) = 0.
$g \text{ is continuous on } \left[0, \frac{\pi}{2}\right]$
cos x and x are continuous everywhere, so their difference is too.
$g(0) = 1 - 0 = 1 > 0, \qquad g\left(\frac{\pi}{2}\right) = 0 - \frac{\pi}{2} < 0$
Opposite signs at the ends, so 0 lies between g(0) and g(π/2).
$g(c) = 0 \text{ for some } c \in \left(0, \frac{\pi}{2}\right), \text{ so } \cos c = c$
The Intermediate Value Theorem. It promises a solution without saying where it is.

Answer

\cos c = c \text{ for some } c \in \left(0, \frac{\pi}{2}\right)

What earns the marks

Rewrites the equation as a zero of one function, such as g(x) = cos x − x.
States that g is continuous on [0, π/2], as the difference of two continuous functions.
Finds the signs at the ends: g(0) = 1 > 0 and g(π/2) = −π/2 < 0.
Concludes by the Intermediate Value Theorem that g(c) = 0 for some c in (0, π/2), so cos c = c.

Check

cos 0.739 ≈ 0.7391, so a solution lies near 0.739, inside the interval.

Where this goes wrong

Checking the signs without saying why the function is continuous, or comparing cos x and x at the ends without making them one function. The theorem is about one continuous function, so rewrite the equation as g(x) = 0 and give the reason g is continuous.