## Calculus: Early Transcendentals 8th Edition

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 $(0,1)$.
$$\cos{\sqrt{x}}=e^x-2$$$$\cos{\sqrt{x}}-e^x+2=0$$ 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 $(0,1)$.