Lesson 8 (09/11/2020)

Lesson 8 (09/11/2020)

Today:

• Definition of continuity
• types of discontinuities
• intermediate value theorem.

Office Hours: Monday, Wednesday, Friday 1:30pm – 2:30pm

Announcements:

• Exam 1 on Wednesday 09/23, 8pm – 9pm
  • \(\hookrightarrow\) Study Guide, instructions, Seating Chart on Brightspace.
• HW 8, 9 Due: Tuesday 09/15
• Quiz 5 (Lesson 7): Tuesday 09/15

Warmup Example:

• \( f(-6) = 5 \)

  \( \lim_{x \to -6} f(x) = 2 \)

  Removable Discontinuity

• \( f(-5) = 3 \)

  \( \lim_{x \to -5} f(x) = \text{DNE} \)

  \( \lim_{x \to -5^-} f(x) = 3 \)

  \( \lim_{x \to -5^+} f(x) = -4 \)

  jump discontinuity

• \( f(3) = \text{Not defined} \)

  \( \lim_{x \to 3} f(x) = +\infty \)

  \( x = 3 \text{ is Vertical Asymptote} \)

  infinite discontinuity

• \( f(1) = -2 \)

  \( \lim_{x \to 1} f(x) = -2 \)

  Continuous at \( x = 1 \)

Definition of Continuity

• there is no hole or jump or vertical Asymptote
• can sketch the graph without lifting hand.

Mathematically, \( f(x) \) is continuous at \( x = a \) if

1. \( f(a) \) is defined
2. \( \lim_{x \to a} f(x) \) exists and is finite
3. \( \lim_{x \to a} f(x) = f(a) \)

if Any of these three conditions are not satisfied the function is discontinuous at \( x = a \)

Examples of some continuous everywhere function:

\[ \underbrace{5x+7, \quad x^2, \quad x^3-2x+1}_{\text{polynomials}}, \quad e^x, \quad \ln x \] \[ \sin x, \quad \cos x \] \[ \left. \frac{x^2+5}{x-1} \right\} \text{Continuous everywhere except when denominator } = 0 \]

In fact, all polynomials, trig functions, exponentials, logarithmic functions, rational functions are all continuous everywhere in their domain.

Example:

\[ f(x) = \frac{x^2 + 2x - 15}{x + 5} \]

Is \( f(x) \) continuous at \( x = 23 \)?

Yes because \( f \) is rational & \( x = 23 \) is in domain.

\( x = -5 \) is the only discontinuity. \( f(-5) \) is not defined.

What is the type of discontinuity? find \( \lim_{x \to -5} f(x) \)

\[ \lim_{x \to -5} \frac{x^2 + 2x - 15}{x + 5} \]

Factorize Numerator \(\to x^2 + 2x - 15 = (x + 5)(x - 3)\)

\[ \lim_{x \to -5} \frac{x^2 + 2x - 15}{x + 5} = \lim_{x \to -5} \frac{\cancel{(x + 5)}(x - 3)}{\cancel{(x + 5)}} = \lim_{x \to -5} x - 3 = -8 \]

\[ f(x) = \frac{x^2 + 2x - 15}{x + 5} = \begin{cases} x - 3 & x \neq -5 \\ \text{Not defined} & x = -5 \end{cases} \]

\( f(x) \) has a Removable discontinuity at \( x=a \) if \( \lim_{x \to a} f(x) \) exists & is finite but not equal to \( f(a) \).

-we can redefine the function just at that one point to make it continuous at \( x=a \)

\[ f(x) = \frac{x^2+2x-15}{x-5} = \begin{cases} x-3 & x \neq -5 \\ \text{Not Defined} & x = -5 \end{cases} \leadsto \text{discontinuous at } x = -5 \]

\[ g(x) = \begin{cases} x-3 & x \neq -5 \\ -8 & x = -5 \end{cases} \leadsto \text{Continuous at } x = -5 \]

eg: \[ f(x) = \frac{1}{x+5} \]

\( f(-5) \) is not defined \( \Rightarrow f(x) \) is discontinuous at \( x = -5 \)

\[ \left. \begin{aligned} \lim_{x \to -5^+} \frac{1}{x+5} &= +\infty \\ \lim_{x \to -5^-} \frac{1}{x+5} &= -\infty \end{aligned} \right\} \quad x = -5 \text{ is a Vertical Asymptote} \]

infinite discontinuity at \( x = -5 \).

eg:

\[ f(x) = \begin{cases} -x^2 + 16 & x \le 0 \\ x + 8 & x > 0 \end{cases} \]

\[ f(0) = 16 \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x + 8 = 8 \] \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 + 16 = 16 \]

Left and Right limit are different

\(\leadsto\) You have a jump discontinuity at \(x = 0\).

Left and Right Continuity

Left continuous at \(x = a\)

1. \(f(a)\) is Defined
2. \(\lim_{x \to a^-} f(x)\) exists & is finite
3. \(\lim_{x \to a^-} f(x) = f(a)\)

Right continuous at \(x = a\)

1. \(f(a)\) is Defined
2. \(\lim_{x \to a^+} f(x)\) exists & is finite
3. \(\lim_{x \to a^+} f(x) = f(a)\)

eg:

\[ f(x) = \begin{cases} x^2 + a & x < 1 \\ b & x = 1 \\ 2x + 3 & x > 1 \end{cases} \]

find \( a, b \) such that \( f(x) \) is continuous at \( x = 1 \).

\( f(1) = b \),

\[ \begin{aligned} \lim_{x \to 1^-} f(x) &= \lim_{x \to 1^-} x^2 + a = 1 + a \\ \lim_{x \to 1^+} f(x) &= \lim_{x \to 1^+} 2x + 3 = 5 \end{aligned} \]

want \( 1 + a = 5 \) for limit to exist \( \Rightarrow a = 4 \)

if \( a = 4 \Rightarrow \lim_{x \to 1} f(x) = 5 \)

want \( f(1) = b = \lim_{x \to 1} f(x) \) for function to be continuous at \( x = 1 \)

so \( b = 5 \).

Intermediate Value Theorem. (IVT)

Suppose \( f(x) \) is continuous on \( [a, b] \)

then \( f(x) \) takes all values between \( f(a) \) & \( f(b) \)

More precisely:

for any \( L \) between \( f(a) \) & \( f(b) \) we can find a \( c \) between \( a \) & \( b \) such that \( f(c) = L \)

Example: \( f(x) = 3x^5 + 2x^3 + 5 \) on \( [-1, 1] \)

\[ f(-1) = -3 - 2 + 5 = 0 \]

\[ f(1) = 3 + 2 + 5 = 10 \]

IVT: for any \( L \) between \( 0 \) & \( 10 \) you can find a number between \( -1 \) & \( 1 \) such that

\[ f(c) = L \]

for eg, \( L = 5 \) between \( 0 \) & \( 10 \)

\[ -1 < 0 < 1 \rightarrow f(0) = 5. \]

Example: Show that \( f(x) = x^4 - 3x - 1 \) has a root in \( (1, 2) \)

(Root: Number such that function value \( = 0 \))

\[ f(1) = 1 - 3 - 1 = -3 < 0 \] \[ f(2) = 16 - 6 - 1 = 9 > 0 \]

IVT is to sketch graph from \( (1, -3) \) to \( (2, 9) \), we have to hit x-axis.

\( \Rightarrow \) there is a root between 1 & 2.