Properties of Continuous Functions

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

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

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 $f(\xi) = c$ .

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

Extreme Value Theorem
Theorem 1..12 Suppose is continuous on . Then there exists at least one number and which attains a maximum and minimum.

Extreme Value Theorem

Theorem 1..12 Suppose $f(x)$

is continuous on $[a,b]$

. Then there exists at least one number $c$

and

which attains a maximum and minimum.

NOTE attains a maximum in iff the following two conditions are satisfied.

Every $x \in [a,b]$ , there exists so that $f(x) \leq M$ .

There exists $\xi$ in such that $f(\xi) = M$ .

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 -axis. Let $f(x) = 3x^{3} - 2x + 5$ . Find so that the value of is positive and the value of is negative. For example,

$\displaystyle f(-2) = -15, \ f1. = 6$
Since is continuous on , no matter how you draw a curve between the points and , 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 $\displaystyle{2\sin{x} - x = 0}$ has a real-valued solution in $\displaystyle{(\frac{\pi}{2},\pi)}$ .

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

$\displaystyle f(\frac{\pi}{2}) = 2\sin{\frac{\pi}{2}} - \frac{\pi}{2} = 2 - \frac{\pi}{2} > 0\$ or $\displaystyle \ f(\pi) = 2\sin{\pi} - \pi = -\pi < 0.$
Since 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) $\displaystyle{\lim_{x \rightarrow 2-}\frac{x-2}{\vert x-2\vert}}$ (b) $\displaystyle{\lim_{x \rightarrow 2+}\frac{x-2}{\vert x-2\vert}}$ (c) $\displaystyle{\lim_{x \rightarrow 1-}\sqrt{\vert x\vert - x}}$ (d) $\displaystyle{\lim_{x \rightarrow 1+}\sqrt{\vert x\vert - x}}$
(e) $\displaystyle{\lim_{x \rightarrow 1-}\frac{\sqrt{x} - 1}{x-1}}$ (f) $\displaystyle{\lim_{x \rightarrow 0-}\frac{\sqrt{x} - 1}{x-1}}$ (g) $\displaystyle{\lim_{x \rightarrow 2+}\frac{\sqrt{x^{2} - 3x + 2}}{x-2}}$

2.

Determine the following functions are continuous at the given point.
(a) $\displaystyle{f(x) = \left\{\begin{array}{ll} x^{2} + 4, & x < 2\\ x^{3}, & x \geq 2 \end{array}\right]; \ x = 2, }$ (b) $\displaystyle{f(x) = \left\{\begin{array}{ll} x^{2} + 4, & x < 2\\ 5, & x = 2\\ x^{3}, & x > 2 \end{array}\right]; \ x = 2, }$
(c) $\displaystyle{f(x) = \left\{\begin{array}{ll} \frac{x^{2} - 1}{x+1}, & x \neq -1\\ -2, & x = -1 \end{array}\right]; \ x = -1, }$
(d) $\displaystyle{f(x) = \left\{\begin{array}{ll} -x^{2}, & x < 0\\ 1 - \sqrt{x}, & x \geq 0 \end{array}\right]; \ x = 0, }$

3.

Define so that the following functions become continuous at ?D
(a) $\displaystyle{f(x) = \frac{x^{2} - 1}{x-1}}$ (b) $\displaystyle{f(x) = \left\{\begin{array}{ll} 2x^{2} + 1, & x < 1\\ 3x^{3}, & x > 1 \end{array}\right]}$

4.

Using the bisection method to approximate the solution of on the interval within the error less than 0.1?D

Exercise B

1.

Find the limit of the followings:
(a) $\displaystyle{\lim_{x \rightarrow 0}\frac{1}{x}}$ (b) $\displaystyle{\lim_{x \rightarrow 0}\frac{1}{x^{2}}}$ (c) $\displaystyle{\lim_{x \rightarrow 0+}\frac{\vert x\vert}{\sqrt{a+x} - \sqrt{a-x}}}$ (d) $\displaystyle{\lim_{x \rightarrow 0-}\frac{x}{\sqrt{1-\cos{x}}}}$
(e) $\displaystyle{\lim_{x \rightarrow \infty}\cos{\frac{1}{x}}}$ (f) $\displaystyle{\lim_{x \rightarrow \infty}\frac{\sin{x}}{x}}$ (g) $\displaystyle{\lim_{x \rightarrow 0+}\cos{\frac{1}{x}}}$

2.

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

3.

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

4.

Find the maximum and minimum of the following functions:
(a) $\displaystyle{f(x) = x^2 - 3x + 1, \ x \in [-2,1]}$ (b) $\displaystyle{f(x) = \frac{1}{x}, \ x \in (0,1]}$
(c) $\displaystyle{f(x) = x^2 - ax, \ x \in [0,2]}$

5.

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

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