## Wednesday, December 20, 2017

### Middle School Math Solutions – Expand Calculator, Perfect Cube

Similarly to the perfect square formula, which we covered in the last post, we get the perfect cube formula. The perfect cube is when we cube a binomial. We use polynomial multiplication and the FOIL method for this formula.

Let’s take a look at it:

(a+b)^3=a^3+3a^2 b+3ab^2+b^3

Here’s the proof:

(a+b)^3=(a+b)(a+b)(a+b)
=(a+b)(a^2+2ab+b^2)
=a(a^2+2ab+b^2 )+b(a^2+2ab+b^2)
=a^3+2a^2 b+ab^2+a^2 b+2ab^2+b^3
=a^3+3a^2 b+3ab^2+b^3

As you can see there are many steps involved, so having this formula will help you solve perfect cube problems quickly.

Let’s see some examples.

Expand (x+1)^3

1.  Apply the formula

a=x,b=1

(x+1)^3=x^3+3∙x^2∙1+3∙x∙1^2+1^3
=x^3+3x^2+3x+1

Expand (2x-2)^3

1. Apply the formula

a=2x,b=-2

(2x-2)^3=(2x)^3+3∙(2x)^2∙(-2)+3∙(2x)∙(-2)^2+(-2)^3
=8x^3-24x^2+24x-8

Expand (s^2+2t)^3

1. Apply the formula

a=s^2,b=2t

(s^2+2t)^3=(s^2 )^3+3∙(s^2 )^2∙(2t)+3∙s^2∙(2t)^2+(2t)^3
=s^6+6s^4 t+12s^2 t^2+8t^3

Not too difficult! Memorizing this formula will help you and become faster at cubing binomials. If you are looking for more practice problems or help, check out Symbolab’s Practice.

Until next time,

Leah

## Wednesday, November 15, 2017

### Middle School Math Solutions – Expand Calculator, Perfect Squares

The perfect square formula is an application of the FOIL method that will help you calculate the square of a binomial quickly.

Let’s take a look at the formula:

(a+b)^2=a^2+2ab+b^2

Here’s the proof:

(a+b)^2=(a+b)(a+b)
=a^2+ab+ab+b^2
=a^2+2ab+b^2

Now, let’s see some examples using the perfect squares formula.

Expand (x+1)^2

1. Apply the formula

a=xb=1

(x+1)^2=x^2+2∙x∙1+1^2
=x^2+2x+1

Expand (5-x)^2

1. Apply the formula

a=5b=-x

(5-x)^2=5^2-2∙5∙(-x)+(-x)^2
=25+10x+x^2

Expand (s^2+4p)^2

1. Apply the formula

a=s^2, b=4p

(s^2+4p)^2=(s^2)^2+2∙s^2∙4p+(4p)^2
=s^4+8s^2 p+16p^2

The more you practice these problems, the faster you will be able to do them. I’ve used this formula so much, that now, I don’t need it since I can do it all mentally. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

## Tuesday, November 7, 2017

### Middle School Math Solutions – Expand Calculator, Two Squares

The difference of two squares is an application of the FOIL method (refer to our blog post on the FOIL method).   The difference of two squares is a number or term squared subtracted from another number or term squared. We get it when we multiply two binomials, where the terms in the binomials are the same, except one of the terms is subtracted instead of being added.

Let’s see this in a formula:
(a+b)(a-b)=a^2-b^2

Here’s the proof using the FOIL method:
(a+b)(a-b)=a^2-ab+ab-b^2
=a^2-b^2

Not too complicated, so let’s see some examples.

Expand (x+2)(x-2)

1. Apply the formula
a=x,b=2

(x+2)(x-2)=x^2-2^2
=x^2-4

This next one is a little more complicated.

Expand (-y+2x)(y+2x)

1. Rewrite the problem
(-y+2x)(y+2x)=(2x-y)(2x+y)
=(2x+y)(2x-y)

2. Apply the formula

a=2x,b=y

(2x+y)(2x-y)=(2x)^2-y^2
=4x^2-y^2

For our last example, we will see an application of the difference of two squares formula.

23∙17

1. Rewrite the numbers
(20+3)(20-3)

2. Apply the formula
a=20,b=3

(20+3)(20-3)=20^2-3^2
=400-9
=391

As you can see, this formula is simple, but very helpful. If you need more help or practice with this formula, check out Symbolab’s Practice.

Until next time,

Leah

## Tuesday, October 31, 2017

### Middle School Math Solutions – Expand Calculator, Polynomial Multiplication

As of right now, you should be familiar with the distributive law and the FOIL method. We will now use both of them and apply them to our next topic, polynomial multiplication.

Recall: A polynomial is an expression of many (poly-) terms (-nomial) that are added and/or subtracted together. The terms include coefficients, variables, and POSITIVE exponents.

Examples: 4x^2 ,2x^3+5x^5 ,\frac{2}{3}x-1

There are a variety of polynomial multiplication problems. They might look intimidating at first, but we will go over how to solve them step by step using the distributive law and FOIL method, so you can see how to do them easily.

4x(3x^2+2x-1)

1. Distribute the 4x

4x(3x^2+2x-1)=4x∙3x^2+4x∙2x+4x∙(-1)
We applied the distributive law here for a polynomial with three terms                  (a(b+c+d)=ab+ac+ad)
2. Simplify

4x(3x^2+2x-1)=12x^3+8x^2-4x
We multiplied the terms.
That one wasn’t bad. Let’s see another one.

(2x+1)(4x+3y+2)

1. Distribute 2x+1

(2x+1)(4x+3y+2)=2x(4x+3y+2)+1(4x+3y+2)
We distributed 2x+1 to make the problem easier for us to solve.
2. Use the distributive law

2x(4x+3y+2)=2x∙4x+2x∙3y+2x∙2

1(4x+3y+2)=4x∙1+3y∙1+2∙1
We plug these values back in
(2x+1)(4x+3y+2)=2x∙4x+2x∙3y+2x∙2+4x∙1+3y∙1+2∙1

3. Simplify

(2x+1)(4x+3y+2)=8x^2+6xy+4x+4x+3y+2
=8x^2+6xy+8x+3y+2

(x+1)(x+2)(x+3)

1. Apply the FOIL method to two of the polynomials

(x+1)(x+2)=x∙x+x∙2+1∙x+1∙2
=x^2+3x+2

2. Plug x^2+3x+2 in for (x+1)(x+2)

(x+1)(x+2)(x+3)=(x^2+3x+2)(x+3)
Now this problem looks like the prior example.
3. Distribute x+3

(x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
I always prefer to distribute the polynomial with the smallest terms, but you can pick to distribute the quadratic. In that case you would have:
(x^2+3x+2)(x+3)=x^2 (x+3)+3x(x+3)+2(x+3)

4. Use the distributive law

(x+1)(x+2)(x+3)=x(x^2+3x+2)+3(x^2+3x+2)
=x∙x^2+x∙3x+x∙2+3∙x^2+3∙3x+3∙2

5. Simplify

(x+1)(x+2)(x+3)=x^3+3x^2+2x+3x^2+9x+6
=x^3+6x^2+11x+6

Not bad! You can see when we break the problem down, step by step, it is a lot easier to solve because we use the basics of expanding (distributive law and FOIL). For more practice problems and help, check out our Practice.

Until next time,

Leah

## Wednesday, October 11, 2017

### Middle School Math Solutions – Expand Calculator, FOIL Method

In our last blog post we covered the distributive law. In this blog post, we will focus on an application of the distributive law, the FOIL method.

The FOIL method is used when multiplying two binomials together. Quick reminder: a binomial is an expression of the sum or difference of two terms.

The FOIL method is when we take the sum of the first two terms multiplied together, the outer terms multiplied together, the inner terms multiplied together, and the last terms multiplied together. This is where we get the acronym FOIL (first, outer, inner last).

Let’s visualize this and see it in a formula:

Let’s see some examples now, using the FOIL method.

(x+3)(x+1)

1.  Use the FOIL method

Remember, we want the sum of the first terms multiplied together, the outer terms multiplied together, the inner terms multiplied together, and the last terms multiplied together.
2.  Simplify

(x+3)(x+1)=x∙x+x∙1+3∙x+3∙1

=x^2+x+3x+3

=x^2+4x+3

Multiply terms and add like terms

Not, too complicated. Let’s see two more examples.

(2x+1)(x-1)

1. Use the FOIL method

2. Simplify

(2x+1)(x-1)=2x^2-2x+x-1

=2x^2-x-1

(3x^2+y)(-x+5y)

1. Use the FOIL method

2. Simplify

(3x^2+y)(-x+5y)=-3x^3+15x^2 y-xy+5y^2

As you can see, multiplying binomials using the FOIL method isn’t hard. The FOIL method is so helpful throughout your math courses, so it is important to memorize and get the hang of it. Once you start practicing, this will become second nature to you. For practice problems or more help on the FOIL method, check out our Practice.

Until next time,

Leah

## Monday, September 25, 2017

### Middle School Math Solutions – Expand Calculator, Distributive Law

The distributive law helps with multiplication problems by breaking down large numbers into smaller numbers. In this blog post, we will talk about the distributive law and how to use it.

The law says that multiplying a number by a group of numbers added together is the same as multiplying each separately. What does this mean? Here we can see it in a formula:

You can see that the multiplication of a has been distributed among the sum of b and c.

Let’s see some examples of how to use the distributive law.

Expand 4(2x+5)

1. Define the values for a, b, c

a=4,   b=2x,   c=5
Refer to the formula to see what these values are.
2. Plug these values into the distributive law formula

4(2x+5)=4∙2x+4∙5

4∙2x+4∙5=8x+20

Expand 5(10-9p)

1. Define the values for a, b, c

a=5,   b=10,   c=-9p

2. Plug these values into the distributive law formula

5(10-9p)=5∙10+5∙(-9p)

5∙10+5∙(-9p)=50-45p
In this last example, we will see another application of the distributive law.

5(106)

Just by looking at this problem, this might be difficult to calculate quickly. So let’s use the distributive law.

1. Make 106 the sum of two numbers

106=100+6

2. Plug this in for 106

5(100+6)

3. Define values a, b, c

a=5,   b=100,   c=6

4. Plug these into the distributive law formula

5(100+6)=5∙100+5∙6

5∙100+5∙6=500+30=530
As you can see, the distributive law is very handy. Therefore, I highly suggest you memorize this formula and become familiar with how to use it. You can do so by doing as many examples as you can. For practice examples and more help, check out Symbolab’s Practice

Until next time,

Leah

## Tuesday, September 19, 2017

### High School Solutions – Functions Calculator, Inverse

Last blog posts, we focused on how to find the domain and range of functions. In this blog post, we will discuss inverses of functions and how to find the inverse of a function.

An inverse of a function f(x), denoted f^(-1)(x), is a function that reverses or undoes f(x). What does that mean? This means that the domain or inputs of a function is the range or output of the function’s inverse and the range or outputs of a function is the domain or inputs of the function’s inverse. If f(x)=y, then f^(-1)(y)=x. Let’s see some pictures to better understand.

It is important to note that some functions have more than one inverse. For example, quadratic equations have two inverses because the negative and positive value for an input goes to the same y value (For x^2+4, when x = 1 and x = -1 , we get y = 5). When you find the inverse of quadratics, you’ll notice you get two inverses, one is a postive square root and the other is a negative square root. We will see an example of this later in the post.

Now, that you’ve got the concept of what the inverse of a function is, we will see the steps on how to find the inverse of a function.

Steps to find the inverse of a function:

1. Replace y for f(x)
2. Solve for x
3. Substitute y = x
4. If you want to check the function, then f(f^(-1) (x))=x  and f^(-1) (f(x))=x

The steps are pretty simple to remember and follow. Now, let’s see some examples.

Find the inverse of f(x)=3x+5

1. Replace y for f(x)

y=3x+5

2. Solve for x

y=3x+5

y-5=3x

\frac{y-5}{3}=x

3. Substitute y = x

\frac{x-5}{3}=y=f^(-1) (x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

f^(-1)(\f(x))=\frac{(3x+5)-5}{3}=\frac{3x}{3}=x

f(f^(-1)(x))=3(\frac{x-5}{3})+5=x-5+5=x

Find the inverse of f(x)=\sqrt{x+3}

1. Replace y for f(x)

y=\sqrt{x+3}

2. Solve for x

y=\sqrt{x+3}

y^2=x+3

y^2-3=x

3. Substitute y = x

x^2-3=y=f^(-1)(x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)

f^(-1)(\f(x))=(\sqrt{x+3})^2-3=x+3-3=x

f(f^(-1)(x))=\sqrt{(x^2-3)+3}=\sqrt{x^2}=x

Last example (

Find the inverse of f(x)=2x^2-2

1. Replace y for f(x)

y=2x^2-2

2. Solve for x

y+2=2x^2

\frac{y+2}{2}=x^2

±\sqrt{\frac{y+2}{2}}=x

3. Substitue y = x

±\sqrt{\frac{x+2}{2}}=y=f^(-1)(x)

4. Check to make sure it is correct (f(f^(-1)(x))=x  and f^(-1)(f(x))=x)
You can do this step on your own. We will skip it to save some time.

As you can see, finding the inverse of a function is pretty simple. It is easy to make algebraic answers, so make sure you check your answer to see if it is correct. With practice, you’ll be able to master this topic easily. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

## Tuesday, August 15, 2017

### High School Solutions – Functions Calculator, Parity (Even or Odd)

In this blog post, we will be discussing the parity of functions and how to find out the parity of a function.

The parity of a function is the attribute of being even, odd, or neither. A function is even, if \f(-x)=\f(x) for all x. A function is odd if \f(-x)=-\f(x) for all x. A function is neither even nor odd, when it satisfies neither of these options.

In order to find the parity of a function, we must see whether the statements about even and odd functions is true or false.

Steps to find the parity of a function:
1.  Find \f(-x)
2.  Find -\f(x)
3.  See if \f(-x)=\f(x), \f(-x)=-\f(x), or neither
Simple enough! Let’s move onto some examples.

Find the parity of \f(x)=x^2+4

1. Find \f(-x)

\f(-x)=(-x)^2+4

\f(-x)=x^2+4

2. Find -\f(x)

-\f(x)=-(x^2+4)=-x^2-4

3.  See if \f(-x)=\f(x),\f(-x)=-\f(x), or neither

\f(x)=x^2+4=\f(-x)

\f(-x)=x^2+4≠-x^2-4=-\f(x)

\f(x) is even.

Find the parity of \f(x)=3x

1. Find f(-x)

\f(-x)=3(-x)=-3x

2. Find -\f(x)

-\f(x)=-(3x)=-3x

3. See if \f(-x)=\f(x),\f(-x)=-\f(x), or neither

\f(-x)=-3x≠3x=\f(x)

\f(-x)=-3x=-\f(x)

\f(x) is odd.

Find the parity of \f(x)=\frac{x+2}{x+1}

1. Find \f(-x)

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

2. Find -\f(x)

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

3. See if \f(-x)=\f(x),\f(-x)=-\f(x), or neither

\f(-x)=\frac{-x+2}{-x+1}≠\frac{x+2}{x+1}=\f(x)

\f(-x)=\frac{-x+2}{-x+1}≠\frac{-x-2}{-x-1}=-\f(x)

\f(x) is neither odd nor even.

As you can see, finding the parity of a function is very simple. Just memorize the definitions of even and odd functions and you are good to go! For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

## Monday, July 31, 2017

### High School Solutions – Functions Calculator, Range (Part II)

Last blog post, we talk about how to find the range of linear, radical, and quadratic functions and what a range is. This week we will learn how to find the range of rational functions, which is trickier.

When it comes to rational functions, there are two ways to find the range depending on if the denominator is a linear expression or if it is a quadratic expression.

Let’s see the steps for each case.

Steps to determining the range of a rational function (denominator is a linear expression):

1.  Find the inverse of the function
• Set the function equal to y, and solve for x
• Substitute y = x at the end
2.  Find the domain of the inverse
• Refer to previous blog post on domains if you need help with this
3.  Write the range
• The domain of the inverse is the range of the function when you substitute y or \f(x) for x
Steps to determining the range of a rational function (denominator is a quadratic expression):

1. Multiply the denominator to both sides of the equation
• Set \f(x)=y
2. Find the discriminant in terms of y
• discriminant= b^2-4ac, given ax^2+bx+c=0
3. Set the discriminant greater than or equal to 0 and solve for y
• Make a table to summarize the results if needed
• Show when the factors of the discriminant and the discriminant are positive, negative, and 0
4. Write the range
• The range is the set of y for which the discriminant is equal to or great than 0
Let’s see an example for when the denominator is a linear expression (click here):

Find the range of \y=\frac{x+3}{x-4}

1. Find the inverse of the function

\y=\frac{x+3}{x-4}

yx-4y-3=x

-4y-3=x(1-y)

x=\frac{-4y-3}{1-y}

y^{-1}=\frac{-4x-3}{1-x}

2. Find the domain of the inverse

1-x=0

x=1

Domain: x<1 or x>1,or (-∞,1)∪(1,∞)

3. Write the range

Range: y<1 or y>1,or (-∞,1)∪(1,∞)

Find the range of \f(x)=\frac{4}{x^2-2x}

1. Multiply the denominator to both sides of the equation

\y=\frac{4}{x^2-2x}

y(x^2-2x)=4

2. Find the discriminant in terms of y

yx^2-2yx-4=0

discriminant= (-2y)^2-4(y)(-4)=4y^2+16y

3. Set the discriminant greater than or equal to 0 and solve for y

4y^2+16y≥0

4y(y+4)≥0

4y is 0 when:y=0                                                                    y+4 is 0 when:y=-4

4y is negative when:y<0                                                        y+4 is negative when:y<-4

4y is positive when:y>0                                                         y+4 is positive when:y>-4

 y<-4 y=-4 -40 y - - - 0 + y+4 - 0 + + + y(y+4) -\:∙\:-\:=+ -\:∙\:0\:=0 -\:∙\:+\:=- 0\:∙\:+\:=0 +\:∙\:+\:=+

4. Write the range

y<-4y=-4y=0y>0

Range:y≤-4 or y≥0, or (-∞,-4)∪(0,∞)

Find the range of y=\frac{x}{x^2+4}

1. Multiply the denominator to both sides of the equation

y(x^2+4)=x

2. Find the discriminant in terms of y

yx^2-x+4y=0

Discriminant= (-1)^2-4(y)(4y)=1-16y^2

3. Set the discriminant greater than or equal to zero and solve for y

1-16y^2≥0

1≥16y^2

\frac{1}{16}≥y^2

y≥\frac{-1}{4}  or  y≤\frac{1}{4}

Note: We did not have to make a table because this was a simpler way to solve for y

4. Write the range

Range: \frac{-1}{4}≤y≤\frac{1}{4}, or [\frac{-1}{4},\frac{1}{4}]

As you can see, finding the range of a function is trickier, especially finding the range of a rational function. It might seem hard and a little scary, but the more practice you get with this, the better you will become. For more help or practice on this topic, visit Symbolab’s Practice.

Until next time,

Leah

## Wednesday, July 5, 2017

### High School Solutions – Functions Calculator, Range (Part I)

Last blog post, we discussed what a domain was and how to find the domain. In this blog post, we will talk about what a range is and how to determine the range of linear, radical, and quadratic functions.

The range of a function is the set of values of the dependent variable (i.e. the y values or output values) for which a function is defined. Another way to think about the range is as the image of the function. The domain is what we can put in the function and the range is what comes out of the function.

Linear functions -

The range of linear functions is always -∞<y<∞ or (-∞,∞) since the function is defined at all the outputs.

Find the range of \f(x)=3x

Range:  -∞<f(x)<∞, or (-∞,∞)

For radical functions there is a simple rule to follow to find the range. Given a radical function, written \f(x)=c\sqrt{ax+b}+k, f(x)≥k is the range.

Find the range of \f(x)=2\sqrt{x+3}-2

We can see here that k = -2

Range:  f(x)≥-2  or (-2,∞)

Finding the range of a quadratic function is a little trickier. When given a quadratic function, we know there is a parabola. The goal is to find the vertex of the parabola and figure out if it is a minimum or a maximum.

Steps to determining the range of a quadratic function:

1. Find the vertex
• x= -\frac{b}{2a},given \f(x)=ax^2+bx+c
• Plug in x into the function to get the y coordinate of the vertex

2. Determine if the vertex is a minimum or a maximum
• If a<0, then the vertex is a maximum
• If a>0, then the vertex is a minimum

3. Write the range
• If the vertex is a maximum, then the range of the function is all the points below and equal to the vertex’s y coordinate
• If the vertex is a minimum, then the range of the function is all the points above and equal to the vertex’s y coordinate

Find the range of \f(x)=x^2+5x+6

1. Find the vertex

x=-\frac{5}{2∙1}=-\frac{5}{2}

f(-\frac{5}{2})=(-\frac{5}{2})^2+5(-\frac{5}{2})+6=-\frac{1}{4}

Vertex is at (-5/2,-1/4)

2. Determine if the vertex is a minimum or a maximum

a=1>0
Vertex is minimum

3. Write the range

Range:\f(x)≥-\frac{1}{4}   or (-\frac{1}{4},∞)

Find the range of \f(x)=-4x^2+2x+4

1. Find the vertex

x=-\frac{2}{2∙-4}=\frac{1}{4}

f(\frac{1}{4})=-4(\frac{1}{4})^2+2(\frac{1}{4})+4=\frac{17}{4}

Vertex is at (\frac{1}{4},\frac{17}{4})

2. Determine if the vertex is a minimum or a maximum

a=-4<0
Vertex is a maximum

3. Write the range

Range:  f(x)≤17/4,  or (-∞,\frac{17}{4})

Finding the range of a function is trickier than finding the domain of a function. We have to think about the outputs instead of the inputs, which can be confusing. The best way to get better at this is to keep practicing and memorizing how to find the range of different functions. Next blog post, we will talk about how to find the range of rational functions. For more help or practice on this topic, visit Symbolab’s Practice.

Until next time,

Leah

## Monday, March 20, 2017

### High School Math Solutions – Functions Calculator, Domain

Determining the domain of a function by looking at its graph is easy to do. However, when you don’t have a graph and have to determine the domain using only the function, it becomes a little harder. In this blog post, we will talk about how to determine the domain of linear, quadratic, radical, and rational functions.

The domain of a function is the set of input or argument values for which the function is real and defined. In simpler terms, the domain is all the values of x that we can plug into the function, keeping the function real and defined. Some examples of values of x that aren’t in the domain are the values of x that cause the function to output an imaginary number or the values of x that cause a zero in the denominator.

Linear and quadratic functions do not have any restrictions on their domain because they are defined and real on all the values of x, unless it’s a piecewise function. That means the domain is -∞<x<∞, which can also be denoted as (-∞,∞).

Find the domain of \f(x)=x^3+5
Domain: -∞<x<∞ or (-∞,∞)
The function is defined and real on all the values of x.

Radical and rational functions, however, do have restrictions on their domains. Radical functions have a restriction on their domain when the expression under the radical is negative. Rational functions have a restriction on their domain whenever the denominator equals zero.

Let’s see the steps to determine the domain.

Determining the domain of radical functions:

1. Set the expression inside the radical greater than or equal to 0
• This is will find the domain, for which the function is real
2. Solve for x

Determining the domain of rational functions:

1. Set the expression in the denominator equal to 0
• This will find the values, for which the function is undefined.
2. Solve for x

Let’s see some examples.

Find the domain of \f(x)=5\sqrt{x^2-9}

1. Set the expression inside the radical greater than or equal to 0

x^2-9≥0

2. Solve for x

x^2-9≥0

x^2≥9

x≥3     or  x≤-3        (*)

Domain:x≥3 or  x≤-3,or (-∞,-3)∪(3,∞)

(*) If \f(x)^2≥a,then \f(x)≥\sqrt{a}  and \f(x)≤-\sqrt{a}

Find the domain of \y=\frac{x}{x^2-6x+8}

1. Set the denominator equal to 0

x^2-6x+8=0

2. Solve for x

x^2-6x+8=0

(x-4)(x-2)=0

x=2 or x=4

Domain:x<2 or  2<x<4 or x>4 ,or (-∞,2)∪(2,4)∪(4,∞)

Note: The domain is the values of x which do not equal 2 or 4, because when x=2 or x=4, the function is undefined (i.e. it causes the denominator to equal 0.

Find the domain of \f(x)=\frac{x+1}{x-1}

1. Set the denominator equal to 0

x-1=0

2. Solve for x

x-1=0

x=1

Domain:x<1 or x>1,or (-∞,1)∪(1,∞)

Finding the domains of functions isn’t too hard. As long as you follow the steps and practice finding the domain multiple times, you will be great at this. For more help or practice on this topic visit Symbolab’s Practice.

Until next time,

Leah