Finding unique solutions

Question

Show that

f(x; k) ≡ cos x − kx = 0 has a unique solution, x0(k), in the interval [0,pi/2]

for all k > 0 .

I don't know where to start

Answer 1

HINT: Draw a sketch of $\cos x$ and $kx\sim mx+c$ in the region $0\leq x\leq\frac{\pi}{2}$.

Observe that the sketch shows a unique intersection where

$$\cos x=kx\Rightarrow \cos x-kx=0.$$

To make this rigorous we might note that the difference between them is decreasing... how might we show that?

Answer 2

Are you allowed to use graphs of functions? If yes, then following argument is easier to visualize:

In interval $[0,\pi/2]$, $f_1(x) = \cos x$ is monotonically decreasing and, $f_2(x) = kx$ is monotonically increasing. Further, $f_1(0) > f_2(0)$ and $f_1(\frac{\pi}{2}) < f_2(\frac{\pi}{2})$, so there is exactly one solution.

Answer 3

Taking derivative w.r.t. $x$, we get $$f^\prime(x:k) = -\sin x -k<0 \mbox{ for all } x \in [0, \pi/2].$$ Hence $f$ is a strictly decreasing function on $[0,\pi/2]$.

Now $f(0) = 1 > 0$ and $f(\pi/2) = -k\pi/2 < 0$.

Moreover, the function $f$ is continuous on the interval $[0,\pi/2]$; so it attains all the values between $f(0) = 1$ and $f(\pi/2) = -k\pi/2$. In particular, it assumes the value $0$ at some point in the open interval $(0, \pi/2)$ at least once.

But as $f$ is a decreasing function, it cannot assume the same value more than once. Hence $f$ has a unique zero in the interval $(0,\pi/2)$.