Evaluate The Function For The Indicated Values Of $x$.$\[ f(x)= \begin{cases} 2x + 1, & X \leq -5 \\ x^2, & -5 \ \textless \ X \ \textless \ 5 \\ 3 - X, & X \geq 5 \end{cases} \\]Calculate:$\[ f(-10) = \square

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Introduction

Piecewise functions are a type of mathematical function that consists of multiple sub-functions, each defined on a specific interval. These functions are commonly used in mathematics, physics, and engineering to model real-world phenomena. In this article, we will evaluate the function for the indicated values of xx using a piecewise function.

The Piecewise Function

The given piecewise function is defined as:

{ f(x)= \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} \}

This function has three sub-functions, each defined on a specific interval:

  • For x5x \leq -5, the function is f(x)=2x+1f(x) = 2x + 1.
  • For 5 \textless x \textless 5-5 \ \textless \ x \ \textless \ 5, the function is f(x)=x2f(x) = x^2.
  • For x5x \geq 5, the function is f(x)=3xf(x) = 3 - x.

Evaluating the Function

To evaluate the function for the indicated values of xx, we need to determine which sub-function is applicable for each value of xx.

Evaluating f(10)f(-10)

For x=10x = -10, we have x5x \leq -5, so the applicable sub-function is f(x)=2x+1f(x) = 2x + 1. Substituting x=10x = -10 into this sub-function, we get:

f(10)=2(10)+1=20+1=19{ f(-10) = 2(-10) + 1 = -20 + 1 = -19 }

Therefore, f(10)=19f(-10) = -19.

Evaluating f(0)f(0)

For x=0x = 0, we have 5 \textless x \textless 5-5 \ \textless \ x \ \textless \ 5, so the applicable sub-function is f(x)=x2f(x) = x^2. Substituting x=0x = 0 into this sub-function, we get:

f(0)=02=0{ f(0) = 0^2 = 0 }

Therefore, f(0)=0f(0) = 0.

Evaluating f(6)f(6)

For x=6x = 6, we have x5x \geq 5, so the applicable sub-function is f(x)=3xf(x) = 3 - x. Substituting x=6x = 6 into this sub-function, we get:

f(6)=36=3{ f(6) = 3 - 6 = -3 }

Therefore, f(6)=3f(6) = -3.

Conclusion

In this article, we evaluated the piecewise function for the indicated values of xx. We determined which sub-function is applicable for each value of xx and substituted the value of xx into the applicable sub-function to find the corresponding value of f(x)f(x). The results are:

  • f(10)=19f(-10) = -19
  • f(0)=0f(0) = 0
  • f(6)=3f(6) = -3

These results demonstrate the importance of carefully evaluating piecewise functions to ensure accurate results.

Discussion

Piecewise functions are a powerful tool in mathematics, allowing us to model complex phenomena with ease. However, they can also be challenging to work with, especially when dealing with multiple sub-functions. In this article, we demonstrated how to evaluate a piecewise function for specific values of xx by determining which sub-function is applicable and substituting the value of xx into the applicable sub-function.

Future Work

In future work, we can explore more advanced applications of piecewise functions, such as using them to model real-world phenomena or to solve complex mathematical problems. We can also investigate the properties of piecewise functions, such as their continuity and differentiability.

References

Glossary

  • Piecewise function: A type of mathematical function that consists of multiple sub-functions, each defined on a specific interval.
  • Sub-function: A function that is defined on a specific interval and is part of a piecewise function.
  • Interval: A range of values that a function is defined on.
  • Applicable sub-function: The sub-function that is applicable for a given value of xx.

Appendix

  • Example 1: Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=10x = -10.
  • Example 2: Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=0x = 0.
  • Example 3: Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=6x = 6.
    Evaluating Piecewise Functions: A Q&A Guide =====================================================

Introduction

In our previous article, we evaluated the piecewise function for the indicated values of xx. However, we received many questions from readers regarding the evaluation of piecewise functions. In this article, we will address some of the most frequently asked questions about evaluating piecewise functions.

Q&A

Q: What is a piecewise function?

A: A piecewise function is a type of mathematical function that consists of multiple sub-functions, each defined on a specific interval.

Q: How do I determine which sub-function is applicable for a given value of xx?

A: To determine which sub-function is applicable for a given value of xx, you need to compare the value of xx with the intervals defined in the piecewise function. If the value of xx falls within an interval, then the corresponding sub-function is applicable.

Q: What if the value of xx falls on the boundary of two intervals?

A: If the value of xx falls on the boundary of two intervals, then you need to check the definition of the piecewise function to determine which sub-function is applicable. In some cases, the piecewise function may be defined differently on the boundary of two intervals.

Q: Can I use a piecewise function to model real-world phenomena?

A: Yes, piecewise functions can be used to model real-world phenomena. For example, you can use a piecewise function to model the cost of a product based on the quantity ordered.

Q: How do I evaluate a piecewise function for a specific value of xx?

A: To evaluate a piecewise function for a specific value of xx, you need to determine which sub-function is applicable for that value of xx and then substitute the value of xx into the applicable sub-function.

Q: What if I get different answers when evaluating a piecewise function using different sub-functions?

A: If you get different answers when evaluating a piecewise function using different sub-functions, then you need to check your work carefully to ensure that you are using the correct sub-function for the given value of xx.

Q: Can I use a piecewise function to solve complex mathematical problems?

A: Yes, piecewise functions can be used to solve complex mathematical problems. For example, you can use a piecewise function to model the behavior of a complex system and then use the piecewise function to solve for the unknown variables.

Examples

Example 1: Evaluating a Piecewise Function

Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=10x = -10.

Solution: Since x=10x = -10 falls within the interval x5x \leq -5, the applicable sub-function is f(x)=2x+1f(x) = 2x + 1. Substituting x=10x = -10 into this sub-function, we get:

f(10)=2(10)+1=20+1=19{ f(-10) = 2(-10) + 1 = -20 + 1 = -19 }

Therefore, f(10)=19f(-10) = -19.

Example 2: Evaluating a Piecewise Function

Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=0x = 0.

Solution: Since x=0x = 0 falls within the interval 5 \textless x \textless 5-5 \ \textless \ x \ \textless \ 5, the applicable sub-function is f(x)=x2f(x) = x^2. Substituting x=0x = 0 into this sub-function, we get:

f(0)=02=0{ f(0) = 0^2 = 0 }

Therefore, f(0)=0f(0) = 0.

Example 3: Evaluating a Piecewise Function

Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=6x = 6.

Solution: Since x=6x = 6 falls within the interval x5x \geq 5, the applicable sub-function is f(x)=3xf(x) = 3 - x. Substituting x=6x = 6 into this sub-function, we get:

f(6)=36=3{ f(6) = 3 - 6 = -3 }

Therefore, f(6)=3f(6) = -3.

Conclusion

In this article, we addressed some of the most frequently asked questions about evaluating piecewise functions. We provided examples and solutions to illustrate the concepts and techniques discussed in the article. We hope that this article has been helpful in clarifying the evaluation of piecewise functions.

Discussion

Piecewise functions are a powerful tool in mathematics, allowing us to model complex phenomena with ease. However, they can also be challenging to work with, especially when dealing with multiple sub-functions. In this article, we demonstrated how to evaluate a piecewise function for specific values of xx by determining which sub-function is applicable and substituting the value of xx into the applicable sub-function.

Future Work

In future work, we can explore more advanced applications of piecewise functions, such as using them to model real-world phenomena or to solve complex mathematical problems. We can also investigate the properties of piecewise functions, such as their continuity and differentiability.

References

Glossary

  • Piecewise function: A type of mathematical function that consists of multiple sub-functions, each defined on a specific interval.
  • Sub-function: A function that is defined on a specific interval and is part of a piecewise function.
  • Interval: A range of values that a function is defined on.
  • Applicable sub-function: The sub-function that is applicable for a given value of xx.

Appendix

  • Example 1: Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=10x = -10.
  • Example 2: Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=0x = 0.
  • Example 3: Evaluate the piecewise function f(x)={2x+1,x5x2,5 \textless x \textless 53x,x5f(x) = \begin{cases} 2x + 1, & x \leq -5 \\ x^2, & -5 \ \textless \ x \ \textless \ 5 \\ 3 - x, & x \geq 5 \end{cases} for x=6x = 6.