Properties of Continuous Functions

Basic properties of continuous functions are Intermediate Value Theorem and Extrem Value Theorem. Intermediate Value Theorem

Theorem 1..11   Let $f(x)$ be a continuous function on $[a,b]$. Suppose that $c$ is a real number satisfying $f(a) < c < f(b)$. Then there exists $\xi \in [a,b]$ so that $f(\xi) = c$.

Figure 1.45: Intermediate Value Theorem

$\xi$ is called ksi or gzai

NOTE Intermediate Value Theorem tells that any continuously varying function takes all values in between. Extreme Value Theorem

Theorem 1..12   Suppose $f(x)$ is continuous on $[a,b]$. Then there exists at least one number $c$ and $d$ which attains a maximum and minimum.

NOTE $f(x)$ attains a maximum in $[a,b]$ iff the following two conditions are satisfied.

1. Every $x \in [a,b]$, there exists $M$ so that $f(x) \leq M$.
2. There exists $\xi$ in $[a,b]$ such that $f(\xi) = M$.

Figure 1.46: Extream Value Theorem

Example 1..29   Show the equation $3x^{3} - 2x + 5 = 0$ has at least one real valued solution.

SOLUTION Note that a equation has a real-valued solution if and only if the graph of the function representing a equation has an intersection with $x$-axis. Let $f(x) = 3x^{3} - 2x + 5$. Find $x$ so that the value of $f(x)$ is positive and the value of $f(x)$ is negative. For example,

$$f(-2) = -15, f1. = 6$$

What is important here is the existence of a slution. So, we do not have to solve the equation. Short cut Before evaluating $f(-2)$, we write $f(x) = 3x^3 - 2x + 5 = x(3x^2 - 2) + 5$. Then, $f(-2) = -2(3(-2)^2 -2) + 5 = -210. + 5 = -15$

Since $f(x)$ is continuous on $[-2,1]$, no matter how you draw a curve between the points $(-2,-15)$ and $(1,6)$, the curve has a point in common. let this point be $\xi$. Then $f(\xi) = 0$ and this $\xi$ is a real-valued solution of $3x^{3} - 2x + 5 = 0$ $\blacksquare$

Exercise 1..29   Show the equation $${2\sin{x} - x = 0}$$ has a real-valued solution in $${(\frac{\pi}{2},\pi)}$$.

If a function changes sign at some point, then the value at the point is 0.

SOLUTION Let $f(x) = 2\sin{x} - x$. Then $f'(x) = 2\cos{x} - 1$ and

$$f(\frac{\pi}{2}) = 2\sin{\frac{\pi}{2}} - \frac{\pi}{2} = 2 - \frac{\pi}{2} > 0$$   or $$f(\pi) = 2\sin{\pi} - \pi = -\pi < 0.$$

Since $f(x)$ is continuous on $(\frac{\pi}{2},\pi)$, by the Intermediate Value Theorem, there exists $\xi \in (\frac{\pi}{2},\pi)$ such that $f(\xi) = 0$ $\blacksquare$

Exercise A

1.

Find the limit of the following functions:

(a) $${\lim_{x \rightarrow 2-}\frac{x-2}{\vert x-2\vert}}$$ (b) $${\lim_{x \rightarrow 2+}\frac{x-2}{\vert x-2\vert}}$$ (c) $${\lim_{x \rightarrow 1-}\sqrt{\vert x\vert - x}}$$ (d) $${\lim_{x \rightarrow 1+}\sqrt{\vert x\vert - x}}$$

(e) $${\lim_{x \rightarrow 1-}\frac{\sqrt{x} - 1}{x-1}}$$ (f) $${\lim_{x \rightarrow 0-}\frac{\sqrt{x} - 1}{x-1}}$$ (g) $${\lim_{x \rightarrow 2+}\frac{\sqrt{x^{2} - 3x + 2}}{x-2}}$$

2.

Determine the following functions are continuous at the given point.

(a) $${f(x) = \left\{\begin{array}{ll} x^{2} + 4, & x < 2\\ x^{3}, & x \geq 2 \end{array}\right]; x = 2, }$$ (b) $${f(x) = \left\{\begin{array}{ll} x^{2} + 4, & x < 2\\ 5, & x = 2\\ x^{3}, & x > 2 \end{array}\right]; x = 2, }$$

(c) $${f(x) = \left\{\begin{array}{ll} \frac{x^{2} - 1}{x+1}, & x \neq -1\\ -2, & x = -1 \end{array}\right]; x = -1, }$$

(d) $${f(x) = \left\{\begin{array}{ll} -x^{2}, & x < 0\\ 1 - \sqrt{x}, & x \geq 0 \end{array}\right]; x = 0, }$$

3.

Define $f(1)$ so that the following functions become continuous at $x = 1$

(a) $${f(x) = \frac{x^{2} - 1}{x-1}}$$ (b) $${f(x) = \left\{\begin{array}{ll} 2x^{2} + 1, & x < 1\\ 3x^{3}, & x > 1 \end{array}\right]}$$

renshu:1-4-4

Using the bisection method to approximate the solution of $f(x) = 7x - 6 = 0$ on the interval $[0,1]$ within the error less than 0.1

Exercise B

1.

Find the limit of the followings:

(a) $${\lim_{x \rightarrow 0}\frac{1}{x}}$$ (b) $${\lim_{x \rightarrow 0}\frac{1}{x^{2}}}$$ (c) $${\lim_{x \rightarrow 0+}\frac{\vert x\vert}{\sqrt{a+x} - \sqrt{a-x}}}$$ (d) $${\lim_{x \rightarrow 0-}\frac{x}{\sqrt{1-\cos{x}}}}$$

(e) $${\lim_{x \rightarrow \infty}\cos{\frac{1}{x}}}$$ (f) $${\lim_{x \rightarrow \infty}\frac{\sin{x}}{x}}$$ (g) $${\lim_{x \rightarrow 0+}\cos{\frac{1}{x}}}$$

enshu:1-4-2

Determine $${f(x) = \left\{\begin{array}{cl} \frac{x^2 - x - 2}{x - 2}, & x \neq 2\\ 3, & x = 2 \end{array} \right.}$$ is continous at $x = 2$.

enshu:1-4-3

Show the function $f(x) = \sqrt{x}$ is continuous on the interval $[0,\infty)$

4.

Find the maximum and minimum of the following functions:

(a) $${f(x) = x^2 - 3x + 1, x \in [-2,1]}$$ (b) $${f(x) = \frac{1}{x}, x \in (0,1]}$$

(c) $${f(x) = x^2 - ax, x \in [0,2]}$$

enshu:1-4-5

Show that the equation $${2\sin{x} - x = 0}$$ has a real solution in the interval $${(\frac{\pi}{2},\pi)}$$