# Recitation 4

Exercise: Locate the discontinuities of the function $y=\frac{1}{1+\sin x}$ and illustrate by graphing.

When $\sin x=-1$, the function is undefined. Thus it is discontinuous at those points, which are $-\frac{\pi}{2}+2k\pi$, where $k$ is any integral number.

Exercise: Show that $f$ is continuous on $(-\infty, \infty)$

$$f(x)=\begin{cases}x^2 & \text{if} x<1 \\ \sqrt{x} & \text{if} x\geq 1\end{cases}$$

We only need to show that $f$ is continuous at $1$. Since $\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^-}x^2=1$, $\lim_{x\rightarrow 1^+}f(x)=\lim_{x\rightarrow 1^+}\sqrt{x}=1$ and $f(1)=1$, we know it is indeed continuous at $1$.

Exercise: For waht value of the constant $c$ is the function $f$ continuous on $(-\infty, \infty)$?

$$f(x)=\begin{cases}cx^2+2x & \text{if} x<2 \\ x^3-cx & \text{if} x\geq 2 \end{cases}$$

We only need to take care of the continuity of the function at $2$. Hence we need $c\cdot 2^2+2\cdot 2=2^3-c\cdot 2$. So $c=\frac{2}{3}$.

Exercise: Show that there is a root of the equation $\cos x = x$ in the interval $(0,1)$.

Let $f(x)=\cos x - x$. Since $f(0)=1>0$ and $f(1)=\cos 1 - 1<0$, by Intermediate Value Theorem $f$ has a root in $(0,1)$.

Exercise: Is there a number that is exactly 1 more than its cube?

Let $x$ be such a number if there is any. Then $x=1+x^3$. Let $f(x)=x^3-x+1$. We know that $f(-2)=-5$ and $f(0)=1$. So the equation has a root in $(-2,0)$.