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Function Notation Express the rule in function notation. (For example, the rule "square, then subtract \(5 "\) is expressed as the function \(f(x)=x^{2}-5 .\) ) Add \(1,\) take the square root, then divide by 6

Short Answer

Expert verified
The function is \( f(x) = \frac{\sqrt{x + 1}}{6} \).

Step by step solution

01

Define the Rule in Words

The rule given is "Add \(1\), take the square root, then divide by \(6\)". We need to convert this rule into function notation.
02

Express the Rule as a Mathematical Expression

Let's express the rule using mathematical operations:1. Add \(1\) to \(x\): \(x + 1\)2. Take the square root: \(\sqrt{x + 1}\)3. Divide by \(6\): \(\frac{\sqrt{x + 1}}{6}\)
03

Rewrite the Expression as a Function

We express the final mathematical expression from Step 2 as a function of \(x\):\[ f(x) = \frac{\sqrt{x + 1}}{6} \]
04

Conclusion: Expressing the Rule in Function Notation

The function that represents the rule "Add \(1\), take the square root, then divide by \(6\)" is \[ f(x) = \frac{\sqrt{x + 1}}{6} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Expressions
Mathematical expressions are like sentences, but instead of words, they use numbers and symbols to convey operations and relationships. In the context of function notation, these expressions allow us to translate word-based rules into a concise mathematical form.
For instance, consider the rule "Add 1, take the square root, then divide by 6." This can be methodically transformed into a mathematical expression through a series of operations.
  • First, we modify the variable by adding 1: \( x + 1 \).
  • Next, we apply the square root to the result: \( \sqrt{x + 1} \).
  • Finally, we divide the whole term by 6: \( \frac{\sqrt{x + 1}}{6} \).
By following these steps, the linguistic rule becomes a precise mathematical expression, which is much easier to work with in further calculations.
Step-by-Step Solution
Breaking down a problem into a step-by-step solution is an effective way to understand and apply function notation. Each step helps to convert a verbal rule into a well-defined function. Let's walk through this process.

**1. Understanding the Rule:**
The first step is to comprehend the operation described in words. For instance, in the exercise, we start by adding 1, then take the square root, and finally divide by 6.

**2. Translating into Mathematical Operations:**
Each part of the rule is separately converted into a mathematical operation:
- Add 1 to a variable \( x \), resulting in \( x + 1 \).
- Apply the square root to \( x + 1 \), yielding \( \sqrt{x + 1} \).
- Divide the result of the square root by 6, giving the final expression \( \frac{\sqrt{x + 1}}{6} \).

**3. Formulate as a Function:**
Once we have the mathematical expression \( \frac{\sqrt{x + 1}}{6} \), it can be captured as a function \( f \) of \( x \), denoted as \( f(x) = \frac{\sqrt{x + 1}}{6} \).

These steps ensure a structured approach, making the transformation from verbal instruction to mathematical function smooth and logical.
Function of x
The notion of a function of \( x \) is foundational in mathematics. It establishes a relationship between inputs and outputs, where an input \( x \) is passed through a series of operations to produce the output \( f(x) \).

In this exercise, the function of \( x \) is built around the operations prescribed by the rule. The mathematical procedure fine-tunes how input \( x \) is altered to deliver an output:
  • First, the input \( x \) is increased by 1.
  • The sum undergoes a square root transformation, adjusting the value further.
  • Lastly, dividing by 6 scales down the result, rendering the final output.
This sequence forms the basis of the function \( f(x) = \frac{\sqrt{x + 1}}{6} \).

Such functions are invaluable because they provide a clear, functional framework to predict output values for any input \( x \). Understanding this connection helps demystify how functions work, illustrating how various sequential processes can be modeled mathematically.

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Most popular questions from this chapter

The table shows the number of DVD players sold in a small electronics store in the years 2003-2013. $$\begin{array}{|c|c|} \hline \text { Year } & \text { DVD players sold } \\ \hline 2003 & 495 \\ 2004 & 513 \\ 2005 & 410 \\ 2006 & 402 \\ 2007 & 520 \\ 2008 & 580 \\ 2009 & 631 \\ 2010 & 719 \\ 2011 & 624 \\ 2012 & 582 \\ 2013 & 635 \\ \hline \end{array}$$ (a) What was the average rate of change of sales between 2003 and 2013 ? (b) What was the average rate of change of sales between 2003 and 2004 ? (c) What was the average rate of change of sales between 2004 and 2005 ? (d) Between which two successive years did DVD player sales increase most quickly? Decrease most quickly?

Graph the function in each of the given viewing rectangles, and select the one that produces the most appropriate graph of the function. \(g(x)=x^{2}-x-20\) (a) \([-2,2]\) by \([-5,5]\) (b) \([-10,10]\) by \([-10,10]\) (c) \([-7,7]\) by \([-25,20]\) (d) \([-10,10]\) by \([-100,100]\)

Coughing When a foreign object that is lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward, causing an increase in pressure in the lungs. At the same time, the trachea contracts, causing the expelled air to move faster and increasing the pressure on the foreign object. According to a mathematical model of coughing, the velocity \(v\) (in \(\mathrm{cm} / \mathrm{s}\) ) of the airstream through an averagesized person's trachea is related to the radius \(r\) of the trachea (in cm) by the function $$v(r)=3.2(1-r) r^{2} \quad \frac{1}{2} \leq r \leq 1$$ Determine the value of \(r\) for which \(v\) is a maximum.

The power produced by a wind turbine depends on the speed of the wind. If a windmill has blades 3 meters long, then the power \(P\) produced by the turbine is modeled by $$P(v)=14.1 v^{3}$$ where \(P\) is measured in watts (W) and \(v\) is measured in meters per second (m/s). Graph the function \(P\) for wind speeds between \(1 \mathrm{m} / \mathrm{s}\) and \(10 \mathrm{m} / \mathrm{s}\). (IMAGE CAN'T COPY).

DISCUSS DISCOVER: Minimizing a Distance When we seek a minimum or maximum value of a function, it is sometimes easier to work with a simpler function instead. (a) Suppose $$g(x)=\sqrt{f(x)}$$ where \(f(x) \geq 0\) for all \(x .\) Explain why the local minima and maxima of \(f\) and \(g\) occur at the same values of \(x .\) (b) Let \(g(x)\) be the distance between the point \((3,0)\) and the point \(\left(x, x^{2}\right)\) on the graph of the parabola \(y=x^{2}\) Express \(g\) as a function of \(x\) (c) Find the minimum value of the function \(g\) that you found in part (b). Use the principle described in part (a) to simplify your work.

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