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

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