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.

First example (click here):

                                                 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

Next example (click here):

                                                     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 (click here):

                                                   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.

First example (click here):

                                                 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.

Next example (click here):

                                                   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.

Last example (click here):

                                             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,∞)

Now let’s see an example when the denominator is a quadratic expression (click here):

                                           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   -4<y<0   y=0   y>0
     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,∞)

We’ll see one more example because it is tricky (click here):

                                               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.

For example (click here):

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

                                                 Range:  -∞<f(x)<∞, or (-∞,∞)

Radical functions -

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.

Here’s an example (click here):

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

We can see here that k = -2

                                                Range:  f(x)≥-2  or (-2,∞)

Quadratic functions -

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

Let’s see an example (click here):

                                             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},∞)

Here’s one more example (click here):

                                                   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 -

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 (-∞,∞).

For example (click here):
                                                     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 -

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.

Here’s an example for radical functions (click here):

                                               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}


Here’s an example for rational functions (click here):

                                            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.

Last example for rational functions (click here):

                                              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

Sunday, February 5, 2017

High School Math Solutions – Derivative Applications Calculator, Tangent Line

We learned in previous posts how to take the derivative of a function. Now, it’s time to see the applications of derivatives.

We use derivatives when we find the equation of a tangent line. A tangent line is a straight line that just touches the curve at a point (on the curve). Tangent lines can help us find the length of the curve and their slopes tell us what the curve looks like and where we can find maximum and minimums. When we take the derivative of the function of the curve at a particular point, we get the slope of the tangent line.



Let’s see how we can use taking the derivative to find the equation of a tangent line.

Steps to find the equation of a tangent line at a point:

1.  Find the tangent point

  • Plug in the value for x into the function to find the y coordinate 

2.  Compute the slope of the function

  • Take the derivative of the function

3.  Compute the slope of the function at the given x coordinate

  • Plug in the value for x into the derivative 

4.  Use the point-slope formula to find the equation of the tangent line

  • y-y_1=m(x-x_1) 
  • Get (x_1, y_1) from Step 1 and get m from Step 3

We’ll now go over some examples.

First example (click here):

                                      Find the tangent line of \f(x)=\sqrt{x^2+1}  at  x=-1

1. Find the tangent point

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

                                                                 (-1,\sqrt{2})

2. Compute the slope of the function

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

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

                                                    \f^' (x)=\frac{x}{\sqrt{x^2+1}

3. Compute the slope of the function at the given x coordinate

                                         \f^' (-1)=\frac{-1}{\sqrt{(-1)^2+1}}=\frac{-1}{\sqrt{2}}

                                                             m=\frac{-1}{\sqrt{2}}

4. Use the point-slope formula to find the equation of the tangent line

                                                  (-1,\sqrt{2})            m=\frac{-1}{\sqrt{2}}

                                                       y-y_1=m(x-x_1)

                                                      y-\sqrt{2}=\frac{-1}{\sqrt{2}}(x-(-1))

                                                      y-\sqrt{2}=\frac{-1}{\sqrt{2}} x-\frac{1}{\sqrt{2}}

                                                        y=\frac{-1}{\sqrt{2}} x+\frac{1}{\sqrt{2}}

Next example (click here):

                                                 Find the tangent line of \f(x)=\frac{1}{x^2}   at (-1,1)
1. Find the tangent point

             We can skip this step because the tangent point is given

2. Compute the slope of the function

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

                                                                     \f^' (x)=-2x^(-3)

                                                                      \f^' (x)=\frac{-2}{x^3}

3. Compute the slope of the function at the x coordinate

                                                                  \f^' (-1)=\frac{-2}{(-1)^3} =2

                                                                               m=2

4. Use the point-slope formula to find the equation of the tangent line

                                                                   (-1,1)          m=2

                                                                      y-1=2(x-(-1))

                                                                           y=2x+3

Last example (click here):

                                             Find the tangent line of \f(x)=x^2+2x+3 at  x=2
1. Find the tangent point

                                                              \f(2)=2^2+2(2)+3=11

2. Compute the slope of the function

                                                                     \f^' (x)=2x+2

3. Compute the slope of the function at the x coordinate

                                                                 \f^' (2)=2(2)+2=6

                                                                           m=6

4. Use the point-slope formula to find the equation of the tangent line

                                                                  (2,11)        m=6

                                                                     y-11=6(x-2)

                                                                         y=6x-1

As you can see, finding the equation of a tangent line of a point on a curve is not too hard. As long as you’ve mastered computing derivatives and the steps to finding the equation of the tangent line, you will be able to solve these problems quick and easily. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

High School Math Solutions – Derivative Applications Calculator, Normal Lines

Last blog post, we talk about using derivatives to compute the tangent lines of functions at certain points. Another application of derivatives is finding the normal line of a function at a certain point.

A normal line is a line that is perpendicular to a tangent line. The slope of the normal line is the negative reciprocal of the slope of the tangent line.



The steps for finding the equation of a normal line is pretty simple, as long as you’ve mastered finding the equation of a tangent line.

Steps for finding the equation of a normal line:

1.  Find the normal point
  • The normal point is the same as the tangent point
2.  Compute the slope of the function at the x coordinate
  • Compute the derivative of the function at the x coordinate 
  • This is the slope of the tangent line
3.  Compute the slope of the perpendicular line at the x coordinate
  • The slope (m) is the negative reciprocal of the slope (m_1) of the tangent line 
  • m=\frac{-1}{m_1} 
4.  Use the point-slope formula to find the equation of the normal line
  • y-y_1=m(x-x_1)


Let’s see some examples.

First example (click here):

                                          Find the normal line of y=x^2-x-1 at x=2

1. Find the normal point

                                                       y=(2)^2-2-1=1

                                                                  (2,1)

2. Compute the slope of the function at the x coordinate

                                                              y^'=2x-1

                                                          y^'=2(2)-1=3

                                                               m_1=3

3. Compute the slope of the perpendicular line at the x coordinate

                                                             m=\frac{-1}{m_1}
                                                                m=\frac{-1}{3}

4. Use the point-slope formula to find the equation of the normal line

                                                      (2,1)       m=\frac{-1}{3}

                                                       y-y_1=m(x-x_1)

                                                          y-1=\frac{-1}{3}(x-2)

                                                           y=\frac{-1}{3} x+\frac{5}{3}

Next example (click here):

                                           Find the normal line of \f(x)=x^4+2e^x  at (0,2)

1. Find the normal point

      It is already given.

2. Compute the slope of the function at the x coordinate

                                                               \f^' (x)=4x^3+2e^x

                                                         \f^' (0)=4(0)^3+2e^0=2

                                                                        m_1=2

3. Compute the slope of the perpendicular line at the x coordinate

                                                                        m=\frac{-1}{2}

4. Use the point-slope formula to find the equation of the normal line

                                                                 (0,2)      m=\frac{-1}{2}

                                                                   y-2=\frac{-1}{2}(x-0)

                                                                     y=\frac{-1}{2} x+2

Last example (click here):

                                              Find the normal line of \f(x)=\frac{1}{x^2}   at x=-1

1. Find the normal point

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

                                                                          (-1,1)

2. Compute the slope of the function at the x coordinate

                                                                  \f^' (x)=\frac{-2}{x^3}

                                                             \f^' (-1)=\frac{-2}{(-1)^3} =2

                                                                        m_1=2

3. Compute the slope of the perpendicular line at the x coordinate

                                                                        m=\frac{-1}{2}

4. Use point-slope formula to find the equation of the normal line

                                                                (-1,1)     m=\frac{-1}{2}

                                                                y-1=\frac{-1}{2}(x-(-1))

                                                                  y=\frac{-1}{2} x+\frac{1}{2}

Finding the equation of a normal line is very similar to finding the equation of a tangent line. Since the steps are similar, make sure you don’t confuse and mix up the definitions of a tangent line and a normal line. For more practice and help with normal lines, check out Symbolab’s Practice.

Until next time,

Leah