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).