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

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

Thursday, December 22, 2016

Advanced Math Solutions – Vector Calculator, Simple Vector Arithmetic

Vectors are used to represent anything that has a direction and magnitude, length. The most popular example of a vector is velocity. Given a car’s velocity of 50 miles per hour going north from the origin, we can draw a vector. In this blog post, we will focus on the simpler aspects of vectors. We won’t talk about how to graph vectors.

Position vectors are vectors that give the position of a point from the origin. The vector is denoted as \vec{v}=<a_1,\:a_2,\:a_3> and starts at point A=(0, 0, 0) and ends at point B=(a_1,\:a_2,\:a_3).

This brings us to how to find a vector given an initial and final point. Given two points A=(a_1,\:a_2,\:a_3) and B=(b_1,\:b_2,\:b_3), the vector \vec{AB}, which goes from point A to B, is \vec{v}=<b_1-a_1,\:b_2-a_2,\:b_3-a_3>.


Adding vectors is very simple

Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and  \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}+\vec{b}=<a_1+b_1,\:a_2+b_2,\:a_3+b_3>

Here’s an example of adding vectors (click here):




Subtracting vectors is just as simple

Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and  \vec{b}=<b_1,\:b_2,\:b_3> , \vec{a}-\vec{b}=<a_1-b_1,\:a_2-b_2,\:a_3-b_3>

Here’s an example subtracting vectors (click here):





Scalar multiplication is used to lengthen or shorten a vector
Given two vectors \vec{a}=<a_1,\:a_2,\:a_3> and any number c, c\vec{a}=<ca_1,\:ca_2\:ca_3>


Here’s an example of scalar multiplication (click here):



Every vector has a magnitude and a direction. The direction is where its arrow is pointed and the magnitude is the length of the vector. If the magnitude of a vector is 1, then we call that vector a unit vector.


Magnitude is denoted as |\vec{a}| or ||\vec{a}||.
We will use ||\vec{a}||, so we don't get confused with absolute values.
||\vec{a}||=\sqrt{a_x^2+a_y^2}


Here’s an example of finding a vector’s magnitude (click here):




A unit vector \hat{u\:}, is a vector with length 1.
\hat{u\:}=\frac{u}{||u||}


Here’s an example of finding the unit vector of a vector (click here):






Here are some properties to memorize about basic vector arithmetic:











The topics we covered in this blog are simple. I recommend practicing a few examples and memorizing the formulas, and you should be good to go. We are going to cover some of the heavier vector topics in next blog.

Until next time,

Leah

Tuesday, December 13, 2016

High School Math Solutions – Inequalities Calculator, Exponential Inequalities

Last post, we talked about how to solve logarithmic inequalities. This post, we will learn how to solve exponential inequalities. The method of solving exponential inequalities is very similar to solving logarithmic inequalities. In these problems, the variable is in the exponent and our goal is to isolate the variable, which means getting it out of the exponent. We will see some these problems in the forms:

                                                                   e^x>a

                                                                   a^x<b

Let’s see how to solve these problems step by step.

1. Use algebraic manipulation to move everything that is not in the exponential expression to one side
2. Isolate the variable by getting rid of the exponential expression

Example:
e^x>a
e^x>e^{\ln(⁡a)}
x>\ln(a)

Example:
a^x<b
a^x<a^{\log_a(b)}
x<\log_a⁡(b)

3. Solve the inequality

These problems are a little simpler than solving logarithmic inequalities. Let’s see how to solve an example step by step.

First example (click here):

                                                     3^{x+1}+1<2

Step 1: Use algebraic manipulation to move everything that is not in the exponential expression to one side

                                                    3^{2x+1}+1<2

                                                       3^{2x+1}<1

Step 2: Isolate the variable by getting rid of the exponential expression

                                                       3^{2x+1}<1

                                                  3^{2x+1}<3^{\log_3(1)}

                                                       2x+1<\log_3⁡(1)

Step 3: Solve the inequality

                                                       2x+1<\log_3(1)

                                                           2x+1<0

                                                            x<\frac{-1}{2}

That was pretty simple. Let’s see some more examples.

Second example (click here):



Last example (click here):



Solving exponential inequalities are simpler than solving logarithmic inequalities. However, it can still get a little tricky when solving these inequalities with more parts. For more help and practice on solving exponential inequalities, visit Symbolab’s practice.

Until next time,

Leah