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Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

$\displaystyle g(x)=2x^2+4x-2$

What is the range of function $\displaystyle g$?

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y>-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y>-4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y<-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y< -4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\leq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\leq-4 \end{Bmatrix}$

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Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

$\displaystyle g(x)=2x^2+4x-2$

What is the range of function $\displaystyle g$?

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y>-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y>-4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y<-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y< -4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\leq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\leq-4 \end{Bmatrix}$

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Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

$\displaystyle g(x)=2x^2+4x-2$

What is the range of function $\displaystyle g$?

Your answer:

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq4 \end{Bmatrix}$

Your answer:

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y>-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y>-4 \end{Bmatrix}$

Your answer:

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y<-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y< -4 \end{Bmatrix}$

Your answer:

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\leq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\leq-4 \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$

Explanation:

This question is testing the concept and understanding of a function's range. It is important to recall that range can be identified graphically or algebraically. Graphically, range contains the y-values that span the image of the function where as domain, contains the x-values of the function. Algebraically, range is known as the output y of a function when input x, is used. In other words, the input values when placed into the function results in the y values creating an (x, y) pair.

For the purpose of Common Core Standards, domain and range fall within the Cluster A of the function and use of function notation concept (CCSS.Math.content.HSF-IF.A).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1. Identify the function and what the question is asking.

$\displaystyle g(x)=2x^2+4x-2$

Find the range of the function.

Step 2. Discuss the options to solve the problem.

I. Graphically plot the function by computer/technology resource. Then interpret the graph.

II. Create a table of (x, y) pairs and plot the points to create the graph. Then interpret the graph.

III. Algebraically find the vertex of the function and using the properties of polynomials determine the range.

For this particular function let's use the third option to find the range.

Solving the quadratic equation to find the range algebraically requires the understanding that the vertex of a parabola represents either the peak (maximum) of a function or the valley (minimum) of the function.

Recall that a quadratic function can be written in the form,

$\displaystyle y=ax^2+bx+c$

and the formula to find the vertex x value is,

$x=-\frac{b}{2a}$ $\displaystyle x=-\frac{b}{2a}$.

Finally to find the y value of the vertex we will substitute the x value found above into the original function. If the quadratic has a positive $\displaystyle a$, then the vertex will occur at the valley (minimum) of the function. If $\displaystyle a$ is negative then the parabola opens down thus creating a peak (maximum) of the function.

Step 3: Use algebraic technique to solve the problem.

$\displaystyle g(x)=2x^2+4x-2$

First, identify the values of the variables.

$\displaystyle a=2, b=4, c=-2$

Next, plug the values into the formula to find the vertex.

$\\ x=-\frac{4}{2(2)} \\ \\x=-\frac{4}{4} \\ \\ x=-1$ $\displaystyle \\ x=-\frac{4}{2(2)} \\ \\x=-\frac{4}{4} \\ \\ x=-1$

Now, substitute the x value found into the function to find the y output value.

$\\g(-1)=2x^2+4x-2 \\g(-1)=2(-1)^2+4(-1)-2 \\g(-1)=2(1)-4-2 \\g(-1)=2-4-2 \\g(-1)=-4$ $\displaystyle \\g(-1)=2x^2+4x-2 \\g(-1)=2(-1)^2+4(-1)-2 \\g(-1)=2(1)-4-2 \\g(-1)=2-4-2 \\g(-1)=-4$

Step 4: Interpret the solution to answer the question.

This question is asking for the range of the function. Using the formula to find the vertex and the properties of quadratics, it was discovered that the vertex is the valley of the function. This means that all other outputs will be greater than the vertex. Therefore, the range will be all real values of y for which y is greater than or equal to negative four. In mathematical terms this is represented by the following.

$\begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$ $\displaystyle \begin{Bmatrix} y\ \epsilon\ \mathbb{R} |y\geq-4 \end{Bmatrix}$

Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

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

What is the domain and range of the function $\displaystyle f$?

$\\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\geq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -2<y<2 \end{Bmatrix}$ $\displaystyle \\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\geq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -2< y< 2 \end{Bmatrix}$

$\\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x> -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\infty<y<\infty \end{Bmatrix}$ $\displaystyle \\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x> -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\infty< y< \infty \end{Bmatrix}$

$\\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\geq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\infty<y<\infty \end{Bmatrix}$ $\displaystyle \\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\geq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\infty< y< \infty \end{Bmatrix}$

$\\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\leq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\infty<y<\infty \end{Bmatrix}$ $\displaystyle \\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\leq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\infty< y< \infty \end{Bmatrix}$

$\\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\geq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\2<y<\infty \end{Bmatrix}$ $\displaystyle \\ \textup{Domain: }\begin{Bmatrix} x\ \epsilon\ \mathbb{R}| x\geq -2 \end{Bmatrix} \\ \\ \textup{Range:} \begin{Bmatrix} y\ \epsilon\ \mathbb{R}| -\2< y< \infty \end{Bmatrix}$

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Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

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

What is the domain and range of the function $\displaystyle f$?

Your answer:

Correct answer:

Explanation:

This question is testing the concept and understanding of a function's domain. It is important to recall that domain can be identified graphically or algebraically. Graphically, range contains the y-values that span the image of the function where as domain, contains the x-values of the function. Algebraically, domain is known as the input values x of a function that then results in the y outputs. In other words, the input values when placed into the function results in the y values creating an (x, y) pair. It is also critical to understand that there are two important restrictions on domain. These restrictions occur when the x variable is in the denominator of a function or under a radical. This is because a fraction does not exist when there is a zero in the denominator; if under the radical is negative it results in imaginary values.

For the purpose of Common Core Standards, domain and range fall within the Cluster A of the function and use of function notation concept (CCSS.Math.content.HSF-IF.A).

Knowing the standard and the concept for which it relates to, we can now do the step-by-step process to solve the problem in question.

Step 1: Identify the function and what the question is asking.

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

Find the domain (x values) and range (y values) of the function. This includes identifying points where x values do not result in a y value. At these points the domain does not exist.

Step 2. Discuss the options to solve the problem.

I. Graphically plot the function by computer/technology resource. Then interpret the graph.

II. Create a table of (x, y) pairs and plot the points to create the graph. Then interpret the graph.

III. Algebraically find the x values that do not exist, these will be areas that are not in the domain.

For this particular function let's use the third option to find the domain.

Any time there is a fraction set the denominator equal to zero and solve for x, this will be a value that x cannot equal since it is not in the domain. Also, anytime there is a radical set the radicand equal to zero and solve for x; this will also be a value x cannot equal as it is also not in the domain.

Fractions:

$y=\frac{1}{a-x} \rightarrow a-x\neq 0\rightarrow a\neq x$ $\displaystyle y=\frac{1}{a-x} \rightarrow a-x\neq 0\rightarrow a\neq x$

Radicals

$y=\sqrt{x-a}\rightarrow x-a\geq0\rightarrow x\geq a$ $\displaystyle y=\sqrt{x-a}\rightarrow x-a\geq0\rightarrow x\geq a$

Step 3: Use algebraic technique to solve the problem.

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

For this particular function there exists a fraction and a radical therefore, the denominator of the fraction needs to be set to zero and solve for x. Also the radicand needs to be set to zero and solve for x.

For the fraction, using algebraic manipulations we get:

$\\ \sqrt{2+x}\neq 0 \\ \\ (\sqrt{2+x})^2\neq0^2 \\ \\2+x\neq0 \\ \\x\neq-2$ $\displaystyle \\ \sqrt{2+x}\neq 0 \\ \\ (\sqrt{2+x})^2\neq0^2 \\ \\2+x\neq0 \\ \\x\neq-2$

For the radicand, using algebraic manipulations we get:

$\\ 2+x\geq0 \\x\geq-2$ $\displaystyle \\ 2+x\geq0 \\x\geq-2$

Interpret the results to identify the domain. Since we found the areas where the domain does not exist we can state the domain as all real values of x except the areas for which we found to not exist. It is important to understand that "all reals" refer to all numbers negative, positive, zero, fractions, and decimal values.

Since the function is linear, the x variable is singular, we know that the range is all real y values.

Therefore, the domain and range solution in mathematical terms is as follows.

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Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

All Common Core: High School - Functions Resources

Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

Flashcards: Domain and Range Relationships: CCSS.Math.Content.HSF-IF.A.1

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