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Oliga
1 day ago
8

Explain how to graph the given piecewise-defined function. Be sure to specify the type of endpoint each piece of the function wi

ll have and why. f(x) = StartLayout enlarged left-brace 1st Row 1st column negative x + 3, 2nd column x less-than 2 2nd row 1st column 3, 2nd column 2 less-than-or-equal-to x less-than 4 3rd Row 1st column 4 minus 2 x, 2nd column x greater-than-or-equal-to 4 EndLayout
Mathematics
2 answers:
Zina [3.9K]1 day ago
4 0

In certain cases, a function necessitates multiple formulas to achieve the desired outcome. An example is the absolute value function \displaystyle f\left(x\right)=|x|f(x)=∣x∣. This function applies to all real numbers and yields results that are non-negative, defining absolute value as the magnitude or modulus of a real number regardless of its sign. It indicates the distance from zero on the number line, requiring all outputs to be zero or greater.

<pwhen inputting="" a="" non-negative="" value="" the="" output="" remains="" unchanged:="">

\displaystyle f\left(x\right)=x\text{ if }x\ge 0f(x)=x if x≥0

<pwhen inputting="" a="" negative="" value="" the="" output="" is="" inverse:="">

\displaystyle f\left(x\right)=-x\text{ if }x<0f(x)=−x if x<0

Due to the need for two distinct operations, the absolute value function qualifies as a piecewise function: a function defined by several formulas for different sections of its domain.

Piecewise functions help describe scenarios where rules or relationships alter as the input crosses specific "boundaries." Business contexts often demonstrate this, such as when the cost per unit of an item decreases past a certain order quantity. The concept of tax brackets also illustrates piecewise functions. For instance, in a basic tax system where earnings up to $10,000 face a 10% tax, additional income incurs a 20% tax rate. Thus, the total tax on an income S would be 0.1S when \displaystyle {S}\leS≤ $10,000 and 1000 + 0.2 (S – $10,000) when S > $10,000.

</pwhen></pwhen>
Zina [3.9K]1 day ago
3 0

Sample Response: To graph the equation f(x) = –x + 3 for x values less than 2, place an open circle at (2, 1) because the first piece’s domain excludes 2. Next, graph a horizontal line at f(x) = 3 for inputs from 2 to 4, marking a closed circle at (2, 3) and an open circle at (4, 3). Finally, sketch f(x) = 4 – 2x for x values starting at 4, ensuring a closed circle at (4, –4) since 4 belongs to the domain of the third segment.

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Elia took \dfrac{5}{11} hour to cycle from her residence to the beach, \dfrac{6}{11} hour to return, and traveled

8\dfrac{2}{11} kilometers from her home to the beach and

8\dfrac{2}{11} kilometers on her way back.

Step-by-step explanation:

1. Let b represent the hours Elia spent cycling to the beach, and p the time taken for the return trip to the park. The combined time for both rides equals 1 hour, represented as

b + p = 1

2. With a constant speed of 18 km/h on her way to the beach, in b hours, she covered 18b kilometers. She then rode back from the beach to the park at a steady speed of 15 km/h, which amounted to 15p kilometers.

Since the distances for both trips are identical, we establish

18b = 15p

3. Solve the two equations:

\left\{\begin{array}{l}b+p=1\\ \\18b=15p\end{array}\right.

Using the first equation

b=1-p

Insert this solution into the second equation

18(1-p)=15p\\ \\18-18p=15p\\ \\18=18p+15p\\ \\33p=18\\ \\p=\dfrac{18}{33}=\dfrac{6}{11}\\ \\b=1-\dfrac{6}{11}=\dfrac{5}{11}

Elia took \dfrac{5}{11} hour to cycle to the beach, \dfrac{6}{11} hour on her return, and covered

18\cdot \dfrac{5}{11}=\dfrac{90}{11}=8\dfrac{2}{11} kilometers going to the beach and

15\cdot \dfrac{6}{11}=\dfrac{90}{11}=8\dfrac{2}{11} kilometers returning to her home.

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