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Rational function

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. Neither the coefficients of the polynomials nor the values taken by the function are necessarily rational.

## Definitions

In the case of one variable, x\,, a function is called a rational function if and only if it can be written in the form

f(x) = \frac{P(x)}{Q(x)}

where P\, and Q\, are polynomial functions in x\, and Q\, is not the zero polynomial. The domain of f\, is the set of all points x\, for which the denominator Q(x)\, is not zero, where one assumes that the fraction is written in its lower degree terms, that is, \textstyle P and \textstyle Q have several factors of the positive degree.

Every polynomial function is a rational function with Q(x) = 1. A function that cannot be written in this form (for example, f(x) = \sin(x)) is not a rational function (but the adjective "irrational" is not generally used for functions, but only for real numbers).

An expression of the form \frac{P(x)}{Q(x)} is called a rational expression. The x need not be a variable. In abstract algebra the x is called an indeterminate.

A rational equation is an equation in which two rational expressions are set equal to each other. These expressions obey the same rules as fractions. The equations can be solved by cross-multiplying. Division by zero is undefined, so that a solution causing formal division by zero is rejected.

## Examples

The rational function f(x) = \frac{x^3-2x}{2(x^2-5)} is not defined at x^2=5 \Leftrightarrow x=\pm \sqrt{5}.

The rational function f(x) = \frac{x^2 + 2}{x^2 + 1} is defined for all real numbers, but not for all complex numbers, since if x were the square root of -1 (i.e. the imaginary unit) or its negative, then formal evaluation would lead to division by zero: f(i) = \frac{i^2 + 2}{i^2 + 1} = \frac{-1 + 2}{-1 + 1} = \frac{1}{0}, which is undefined.

The rational function f(x) = \frac{x^3-2x}{2(x^2-5)}, as x approaches infinity, is asymptotic to \frac{x}{2}.

A constant function such as f(x) = Ï€ is a rational function since constants are polynomials. Note that the function itself is rational, even though f(x) is irrational for all x.

The rational function f(x) = \frac{x}{x} is equal to 1 for all x except 0, where there is a removable discontinuity. The sum, product, or quotient (excepting division by the zero polynomial) of two rational functions is itself a rational function: however, the process of reduction to standard form may inadvertently result in the removing of such discontinuities unless care is taken.

## Taylor series

The coefficients of a Taylor series of any rational function satisfy a linear recurrence relation, which can be found by setting the rational function equal to its Taylor series and collecting like terms.

For example,

\frac{1}{x^2 - x + 2} = \sum_{k=0}^{\infty} a_k x^k.

Multiplying through by the denominator and distributing,

1 = (x^2 - x + 2) \sum_{k=0}^{\infty} a_k x^k
1 = \sum_{k=0}^{\infty} a_k x^{k+2} - \sum_{k=0}^{\infty} a_k x^{k+1} + 2\sum_{k=0}^{\infty} a_k x^k.

After adjusting the indices of the sums to get the same powers of x, we get

1 = \sum_{k=2}^{\infty} a_{k-2} x^k - \sum_{k=1}^{\infty} a_{k-1} x^k + 2\sum_{k=0}^{\infty} a_k x^k.

Combining like terms gives

1 = 2a_0 + (2a_1 - a_0)x + \sum_{k=2}^{\infty} (a_{k-2} - a_{k-1} + 2a_k) x^k.

Since this holds true for all x in the radius of convergence of the original Taylor series, we can compute as follows. Since the constant term on the left must equal the constant term on the right it follows that

a_0 = \frac{1}{2}.

Then, since there are no powers of x on the left, all of the coefficients on the right must be zero, from which it follows that

a_1 = \frac{1}{4}
a_{k} = \frac{1}{2} (a_{k-1} - a_{k-2})\quad for\ k \ge 2.

Conversely, any sequence that satisfies a linear recurrence determines a rational function when used as the coefficients of a Taylor series. This is useful in solving such recurrences, since by using partial fraction decomposition we can write any rational function as a sum of factors of the form 1 / (ax + b) and expand these as geometric series, giving an explicit formula for the Taylor coefficients; this is the method of generating functions.

## Complex analysis

In complex analysis, a rational function

f(z) = \frac{P(z)}{Q(z)}

is the ratio of two polynomials with complex coefficients, where Q is not the zero polynomial and P and Q have no common factor (this avoids f taking the indeterminate value 0/0). The domain and range of f are usually taken to be the Riemann sphere, which avoids any need for special treatment at the poles of the function (where Q(z) is 0).

The degree of a rational function is the maximum of the degrees of its constituent polynomials P and Q. If the degree of f is d then the equation

f(z) = w \,

has d distinct solutions in z except for certain values of w, called critical values, where two or more solutions coincide. f can therefore be thought of as a d-fold covering of the w-sphere by the z-sphere.

Rational functions with degree 1 are called

Formula calculator

A formula calculator is a software calculator that can perform a calculation in two steps:

1. Enter the calculation by typing it in from the keyboard.

2. Press a single button or key to see the final result.

This is unlike button-operated calculators, such as the Windows Calculator or the Mac os calculator, which require the user to perform one step for each operation, by pressing buttons to calculate all the intermediate values, before the final result is shown.

In this context, a formula is also known as an expression, and so formula calculators may be called expression calculators. Also in this context, calculation is known as evaluation, and so they may be called formula evaluators, rather than calculators.

## How they work

Formulas as they are commonly written use infix notation for binary operators, such as addition, multiplication, division and subtraction. This notation also uses:

• Parentheses to enclose parts of a formula that must be calculated first.
• In the absence of parentheses, operator precedence, so that higher precedence operators, such as multiplication, must be applied before lower precedence operators, such as addition. For example, in 2 + 3*4, the multiplication, 3*4, is done first.
• Among operators with the same precedence, associativity, so that the left-most operator must be applied first. For example, in 2 - 3 + 4, the subtraction, 2 - 3, is done first.

Also, formulas may contain:

• Non-commutative operators that must be applied to numbers in the correct order, such as subtraction and division.
• The same symbol used for more than one purpose, such as - for negative numbers and subtraction.

Once a formula is entered, a formula calculator follows the above rules to produce the final result by automatically:

• Analysing the formula and breaking it down into its constituent parts, such as operators, numbers and parentheses.
• Finding both operands of each binary operator.
• Working out the values of these operands.
• Applying the operator to these values, in the correct order so as to allow for non-commutative operators.
• Evaluating the parts of a formula in parentheses first.
• Taking operator precedence and associativity into account.
• Distinguishing between different uses of the same symbol.

## How to enter formulas

### Operators

Formulas printed in many text books use juxtaposition, underline and superscripts for multiplication, division and exponentiation respectively. Also, some operations, such as square root, are represented by special symbols that are not usually available on a computer keyboard. For example, see the formulas in Amortization calculator, Heron's formula and Law of cosines.

### Multiplication

In many software tools, including spreadsheets and programming languages, the asterisk, *, is used for multiplication. However, it is also possible to use juxtaposition. For example:

2cos(3)

means 2 multiplied by cos(3).

In calculators that donâ€™t allow juxtaposition, the asterisk (and possibly x rather than, or as well as, the asterisk), is used, and the calculation should be entered as:

2*cos(3)

Also, a period is sometimes used for multiplication, as in:

2.cos(3)

Because the period is also used as the decimal point in numbers, so that the â€œ2.â€� in the above would be interpreted as 2.0, the period is not used for multiplication in a formula calculator, and this calculation should be entered using a different symbol, as above.

### Division

Printed formulas often use a horizontal line for division, but in a formula calculator that uses only keyboard symbols, division is entered using the forward slash, /. When there is a calculation above or below the line, this should be done first, and so it should be enclosed in parentheses when typed in. For example,

2 + 3 â€”â€”â€”â€”â€” 4 - 5

should be entered as

(2 + 3)/(4 - 5)

Also, the symbol Ã· is often used for division, as in

2 Ã· 3

This symbol is not available on most computer keyboards, so this division operation is entered using the forward slash, as above.

### Exponentiation

Exponentiation, or raising to a power, is often represented using a superscript. For example:

2.452

means 2.45 squared.

With the limitations of a computer keyboard, in some software packages, such as Microsoft Excel, this is entered using the caret, ^:

2.45^2

but two asterisks are also used:

2.45**2

### Exponentiation and functions

When using functions, the superscript is sometimes placed immediately after the function name. For example, it is common to write the trigonometric version of Pythagorasâ€™ Theorem (List of trigonometric identities) as:

sin2(x) + cos2(x) = 1

In this identity,

sin2(x)

means the square of sin(x), and is the same as:

sin(x)2

So, for x = 3.25, it could be entered into a formula calculator that uses the caret for exponentiation as:

sin(3.25)^2

### Square roots

Square roots are often specified using the âˆš symbol, but with the limitations of a keyboard this is can be entered by using exponentiation. For example, the square root of 2:

âˆš2

could be entered as:

2^(1/2)

but two asterisks are also used:

2**(1/2)

The parentheses specify that the division should be done first.

### Other roots

All roots can be specified in this way. For example, the cube root of 2 can be entered as:

2^(1/3)

### Heronâ€™s formula example

An example that illustrates these features is Heronâ€™s formula.

One version of the formula, for a triangle with sides of length a, b and c, is equivalent to, using symbols that are available on a keyboard:

1/2*(a^2*c^2 - (a^2 + c^2 - b^2)/2)^0.5

For a = 2.5, b = 3.6 and c = 1.9, it could be entered into a formula calculator as:

1/2*(2.5^2*1.9^2 - (2.5^2 + 1.9^2 - 3.6^2)/2)^0.5

## Types of calculator

The formula calculator concept can be applied to all types of calculator, including arithmetic,

Question:list all numbers which the rational expression is unidentified. 3 ------ a - 8 or 3 / a - 8 is it: 24, -8, 11, 0, or 8 or none of these. if none of these what is the answer? this is a multi choice question. List all the numbers which the rational expression is undefined. problem is 3/a-8 choices are A) -8 B) 0 C) 24 D) 11 E) 8 or F) none of these.

Answers:I'm sure you mean UNDEFINED as "unidentified" is not a math term at all. An algebraic expression is undefined wherever a zero would result in the denominator. Therefore, 3/(a-8) is undefined when a-8=0, which means the expression is undefined when a=8. that's it! ;)

Question:1) Find all numbers for which the rational expression is undefined . 4/v - 8 2.) Divide and simplify c + 6/c - 7 / 2c + 12/c - 1 3.) Solve z - 1/z - 4 = 3/z - 4 4.) Find the LCM of 7(x - 2) and 14(x - 2) 5.)Solve x/x + 7 - 7/x - 7 = x^2 + 49/x^2 - 49 6.) Simplify by removing factors of 1. 48w^8 y^9/18w^4 y^3 7.) Add and simplify if possible 8t/t^2 - 9 + t/t - 3 8.) Find the LCM v^3 + 8v^2 + 16v, v^2 - 8v 9.) Divide and simplify x^2 + 9x/x^2 + 3x - 54 / x x + 6 10.) Simplify by removing factors of 1. t^2 - 36/(t + 6)^2 11.) Subtract and simplify if possible 1/3s^2 - 3s - 5/3s - 3 12.) Simplify by removing factors of 1. y^2 - 6y + 5/y^2 + 5y - 6 13.) Multiply and simplify 3x^3/7x^2 * 49x^4/9x 14.)Subtract and simplify if possible. 19/b - 5 - 19/5 - b 15.)Add and simplify if possible 3/4z^3 + 1/6z^2 16.)Solve x + 8/x = -9 17.) Solve 8/z + 6 = 7/z + 5 18.) Add and simplify if possible 4b/5b - 10 + 9b/10b - 20 Someone please help me with these problems I am having a test on these and I am not sure on how to do them so anything anyone can do and show work so I will know how you got your answer I would greatly appreciate.

Answers:Here are a few. 1) Find all numbers for which the rational expression is undefined . 4/v - 8 You can't divide by zero. Set the denominator equal to zero to find where the expression is undefined. The denominator is just v, so v = 0 is where the expression is undefined. If you meant 4/(v-8), then the denominator is v-8. v-8 = 0 means v = 8, so the expression is undefined when v = 8. I hope you see the DIFFERENCE between 4/v - 8 and 4/(v-8). 2.) Divide and simplify c + 6/c - 7 / 2c + 12/c - 1 With no parentheses, your expression is difficult to interpret. What are the denominators? numerators? 3.) Solve z - 1/z - 4 = 3/z - 4 Again, you don't have parentheses, but I THINK you mean (z-1)/(z-4) = 3/(z-4) ....... but I am not 100% sure. You could also mean z - 1/(z-4) = 3/(z-4). Or possibly something else. I'll assume you mean (z-1)/(z-4) = 3/(z-4). Since the denominators are equal, you can just drop them. z - 1 = 3 Add 1 to each side. z = 4 Then go back to the original equation to see if this solution makes sense. If you plug in z=4, then you will be dividing by zero. Therefore, there is no solution. 4.) Find the LCM of 7(x - 2) and 14(x - 2) The LCM is 14(x-2). 5.)Solve x/x + 7 - 7/x - 7 = x^2 + 49/x^2 - 49 Still no parentheses, but let's assume you mean x/(x+7) - 7/(x-7) = (x^2+49)/(x^2-49) Note that x^2-49 = (x-7)(x+7). Multiply both sides by (x-7)(x+7). x(x-7) - 7(x+7) = x^2 +49 x^2 -7x -7x -49 = x^2 +49 -14x - 49 = 49 -14x = 98 x = -7 Does this solution work in the original equation? I'll let you determine that. I'll also let you attempt the rest.

Question:How do I simplify -4x / 3xy^2 Also what are some rules of simplifiying these kind of problems? Im having trouble doing it when there are fractions and exponents, and multiple variables, etc/

Answers:-4x / 3xy^2 = -4 / 3y^2 [Because the x in the denominator (bottom bit) cancells with the x in the numerator (top bit) But that's all that can be done with it. As for the rules if you google for them you'll find them Here's a powerpoint version thats pretty good: http://www.ltscotland.org.uk/Images/therulesofindices_tcm4-123386.ppt It doesn't give the rule for fractional exponents because thats in another presentation which wasn't listed and I'm too lazy to try and find it but this rule might help: (a^m)^n = a^(mn) is used to help evaluate terms with fractional exponents a^(p/q) means the qth root of a^p So a^(p/q) means (a^p)^(1/q) but to simplify stuff its probably better to think of it as meaning [a^(1/q)]^p because this is easier to evaluate without having to use a calculator. When you have a fraction raised to an exponent the exponent refers to both the numerator and the denominator of the fraction For example (8/27)^(2/3) = [8^(2/3)] / [27^(2/3)] = [8^(1/3)]^2 / [27^(1/3)]^2 = 2^2 / 3^2 = 4 / 9 Not sure if that'll help ... hope it does.

Question:For numbers 1-3, Find all the values that make the rational expression(s) undefined. 1. (m + 4) / (m - 6) = 2. (r + 6) / (r^2 - r - 6) = 3. (9y + 8) / (y) = For numbers 4-10, Write each rational expression in lowest terms. State all the values for which the denominator of the original problem would be undefined. 4. (36m^4n^3) / (24m^2n^5) = 5. ((x + 4)(x - 3)) / ((x + 5)(x + 4)) = 6. (3z^2 + z) / (18z + 6) = 7. (5x - 25) / (x^2 - 25) = 8. (3x^2 - 3) / (6x - 6) = 9. (y^2 + 2y - 3) / (y^2 - 3y + 2) = 10. (m - 3) / (3 - m) = Try to show all work, too! Pleeease :) Thanks.

Answers:Some guidance. Numbers 1-3... for the expressions to be undefined the denominator must be zero. Numbers 4-10... factorise each bracketed term and then see if you can cancel any like terms. Eg in question 5 you have (x+4) in both the denominator and numerator. So these cancel each other out resulting in (x-3)/(x+5). As I said, for the others you will have to factorise first before cancelling.