Question Statement

Find the points of discontinuity for the function:

$f (x) = \frac{1}{x ^{2} - 9 x + 8}$

Background and Explanation

Rational functions are discontinuous wherever the denominator equals zero, since division by zero is undefined. To find points of discontinuity, we factor the denominator polynomial and identify the values of $x$ that make it equal to zero.

Solution

We begin with the given function:

$f (x) = \frac{1}{x ^{2} - 9 x + 8}$

To find where the function is undefined, we factor the quadratic expression in the denominator. We split the middle term $- 9 x$ into $- 8 x$ and $- x$ :

$f (x) = \frac{1}{x ^{2} - 9 x + 8} = \frac{1}{x ^{2} - 8 x - x + 8}$

Now we factor by grouping. From the first two terms we factor out $x$ , and from the last two terms we factor out $- 1$ :

$= \frac{1}{x ( x - 8 ) - 1 ( x - 8 )}$

We can see that $(x - 8)$ is a common factor in both terms:

$f (x) = \frac{1}{( x - 8 ) ( x - 1 )}$

The function is undefined when the denominator equals zero. Setting each factor equal to zero:

$(x - 8) (x - 1) = 0$

This gives us: $x = 8 or x = 1$

Clearly $f (x)$ is not defined at $x = 1$ and $x = 8$ . Therefore, $f (x)$ is discontinuous at $x = 1$ and $x = 8$ .

Hence, the points of discontinuity are $x = 1$ and $x = 8$ .

Key Formulas or Methods Used

Factorization by grouping: Splitting the middle term of a quadratic $a x^{2} + b x + c$ to create factorable groups
Condition for discontinuity: A rational function $f (x) = \frac{P ( x )}{Q ( x )}$ is discontinuous at all values of $x$ where $Q (x) = 0$
Zero product property: If $ab = 0$ , then either $a = 0$ or $b = 0$

Summary of Steps

Identify the denominator: Recognize that discontinuities occur where $x^{2} - 9 x + 8 = 0$
Factor the quadratic: Split $- 9 x$ into $- 8 x - x$ to get $x^{2} - 8 x - x + 8$
Group and factor: Factor out $x$ from the first pair and $- 1$ from the second pair to obtain $x (x - 8) - 1 (x - 8)$
Extract common factor: Factor out $(x - 8)$ to get the final form $(x - 8) (x - 1)$
Find zeros: Set $(x - 8) (x - 1) = 0$ and solve to find $x = 1$ and $x = 8$
State discontinuities: Conclude that the function is discontinuous at $x = 1$ and $x = 8$

Solution

We begin with the given function:

f (x) = \frac{1}{x ^{2} - 9 x + 8}

To find where the function is undefined, we factor the quadratic expression in the denominator. We split the middle term

- 9 x

into

- 8 x

and

- x

f (x) = \frac{1}{x ^{2} - 9 x + 8} = \frac{1}{x ^{2} - 8 x - x + 8}

Now we factor by grouping. From the first two terms we factor out

x

, and from the last two terms we factor out

- 1

= \frac{1}{x ( x - 8 ) - 1 ( x - 8 )}

We can see that

(x - 8)

is a common factor in both terms:

f (x) = \frac{1}{( x - 8 ) ( x - 1 )}

The function is undefined when the denominator equals zero. Setting each factor equal to zero:

(x - 8) (x - 1) = 0

This gives us:

x = 8 or x = 1

Clearly

f (x)

is not defined at

x = 1

and

x = 8

. Therefore,

f (x)

is discontinuous at

x = 1

and

x = 8

Hence, the points of discontinuity are

x = 1

and

x = 8

Summary of Steps

Identify the denominator: Recognize that discontinuities occur where

x^{2} - 9 x + 8 = 0

Factor the quadratic: Split

- 9 x

into

- 8 x - x

to get

x^{2} - 8 x - x + 8

Group and factor: Factor out

x

from the first pair and

- 1

from the second pair to obtain

x (x - 8) - 1 (x - 8)

Extract common factor: Factor out

(x - 8)

to get the final form

(x - 8) (x - 1)

Find zeros: Set

(x - 8) (x - 1) = 0

and solve to find

x = 1

and

x = 8

State discontinuities: Conclude that the function is discontinuous at

x = 1

and

x = 8

2.2 Q-3

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.2 Q-3

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps