Answer
$f(0)f(1)<0,f$ continuous$\Rightarrow \exists x_0\in(0,1),f(x_0)=0$
Work Step by Step
We are given the function:
$f(x)=\cos x-\cos^{-1} x$
First we determine the domain of $f(x)$.
The domain of $\cos x$ is $(-\infty,\infty)$, while the domain of $\cos^{-1} x$ is $[-1,1]$,therefore the domain of $f$ is:
$(-\infty,\infty)\cap [-1,1]=[-1,1]$
Compute $f(-1)$ and $f(1)$:
$f(-1)=\cos (-1)-\cos^{-1}(-1)\approx -2.6$
$f(1)=\cos 1-\cos^{-1}1\approx 0.54$
As $f(x)$ is continuous on $(-1,1)$ and $f(-1)$ and $f(1)$ have opposite signs, it means that the function has a zero between -1 and 1.
Compute $f(0)$:
$f(-1)=\cos 1-\cos^{-1}1\approx -0.57$
We can tell that there is a zero between 0 and 1.
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