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

Tuesday, January 24, 2017

Advanced Math Solutions – Vector Calculator, Advanced Vectors

In the last blog, we covered some of the simpler vector topics. This week, we will go into some of the heavier vector topics. This includes dot product, cross product, and projection. So let’s hop into it.

























Let’s see some examples using the dot product . . .
First example (click here):



Next example (click here):




Now, we will learn about cross product. In multiplication, we often see 1×1 or 1∙1, which both equal the same thing. However, these symbols are very different when talking about vectors. It is important to not interchange the symbols.
















While finding the cross product of two vectors, it is important to know what direction the cross product of the two vectors will point. This is where the right hand rule comes into play.








We’ll go over some properties of cross products.














Now that you have a brief overview on cross products, here’s an example on how to find the cross product of two vectors (click here):




When finding the cross product of two vectors, I find it easier to make a matrix and derive the answer from the matrix, rather than memorizing the formula for the answer.


Onto the last topic, projection . . .












Simple enough, let’s see an example (click here):




In this blog post, we learned about three heavy vector topics. The key to being successful in these topics is to memorize the formulas. If you do that, this will be a piece of cake. Use the definitions, properties, and facts to familiarize yourself with how to find if two vectors are orthogonal or parallel; these tend to be teachers’ favorite questions.

Until next time,

Leah