#### Answer

1) Prove that the function $f(x)=\cos{\sqrt{x}}-e^x+2$ is continuous on its domain.
2) Calculate $f(0)$ and $f(1)$
3) Show that $0\in(f(0),f(1))$. From that, according to the Intermediate Value Theorem, we can show that there must be a number $a$ in the interval $(0,1)$ so that $f(a)=0$. That means there is a root of the equation in interval $(0,1)$.

#### Work Step by Step

$$\cos{\sqrt{x}}=e^x-2$$$$\cos{\sqrt{x}}-e^x+2=0$$ interval $(0,1)$
*STRATEGY:
1) Prove that the function $f(x)=\cos{\sqrt{x}}-e^x+2$ is continuous on its domain.
2) Calculate $f(0)$ and $f(1)$
3) Show that $0\in(f(0),f(1))$. From that, according to the Intermediate Value Theorem, we can show that there must be a number $a$ in the interval $(0,1)$ so that $f(a)=0$. That means there is a root of the equation in interval $(0,1)$.
1) Function $y=\cos{\sqrt x}$ is continuous on its domain, which is $x\ge0$
Function $y=e^x$ is continuous on its domain, which is $R$
Therefore, function $f(x)=\cos{\sqrt{x}}-e^x+2$ is continuous on its domain, which is $x\ge0$.
The interval $(0,1)$ lies in the domain of $f(x)$.
2) Calculate:
$f(0)=\cos{\sqrt{0}}-e^0+2=1-1+2=2$
$f(1)=\cos{\sqrt{1}}-e^1+2\approx0.54-e+2\approx-0.178$
3) Since $f(0)\gt0$ and $f(1)\lt0$, $0$ lies in the interval $(f(0), f(1))$.
According to the Intermediate Value Theorem, since $f(x)$ is continuous on $(0,1)$, there must a number $a$ in $(0,1)$ that $f(a)=0$.
That means there must be a root of the equation in the interval $(0,1)$.