Determine { F(5) $}$ For ${ F(x) = \begin{cases} x^3, & X \ \textless \ -3 \ 2x^2 - 9, & -3 \leq X \ \textless \ 4 \ 5x + 4, & X \geq 4 \end{cases} }$A. 11 B. 29 C. 41 D. 125

Introduction

In mathematics, a piecewise function is a function that is defined by multiple sub-functions, each applied to a specific interval of the domain. These sub-functions are often referred to as "pieces" of the function, and they are used to define the function's behavior over different intervals. In this article, we will explore how to determine the value of a piecewise function at a specific point.

Understanding Piecewise Functions

A piecewise function is defined as:

$$f(x) = \begin{cases} f_1(x), & x \in I_1 \\ f_2(x), & x \in I_2 \\ \vdots \\ f_n(x), & x \in I_n \end{cases}$$

where $f_1(x), f_2(x), \ldots, f_n(x)$ are the sub-functions, and $I_1, I_2, \ldots, I_n$ are the intervals over which each sub-function is defined.

The Given Piecewise Function

The given piecewise function is:

$$f(x) = \begin{cases} x^3, & x < -3 \\ 2x^2 - 9, & -3 \leq x < 4 \\ 5x + 4, & x \geq 4 \end{cases}$$

Determining the Value of $f(5)$

To determine the value of $f(5)$, we need to find the sub-function that is defined over the interval containing $x = 5$. In this case, the interval is $x \geq 4$, so we will use the sub-function $f(x) = 5x + 4$.

Calculating the Value of $f(5)$

To calculate the value of $f(5)$, we substitute $x = 5$ into the sub-function $f(x) = 5x + 4$:

$$f(5) = 5(5) + 4 = 25 + 4 = 29$$

Conclusion

In this article, we have determined the value of a piecewise function at a specific point. We have seen how to identify the sub-function that is defined over the interval containing the point, and how to calculate the value of the function at that point. The value of $f(5)$ is $\boxed{29}$.

Additional Examples

To further illustrate the concept of determining the value of a piecewise function, let's consider a few additional examples.

Example 1

Determine the value of $f(-2)$ for the piecewise function:

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

Solution

To determine the value of $f(-2)$, we need to find the sub-function that is defined over the interval containing $x = -2$. In this case, the interval is $x < -1$, so we will use the sub-function $f(x) = x^2$.

To calculate the value of $f(-2)$, we substitute $x = -2$ into the sub-function $f(x) = x^2$:

$$f(-2) = (-2)^2 = 4$$

Example 2

Determine the value of $f(0)$ for the piecewise function:

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

Solution

To determine the value of $f(0)$, we need to find the sub-function that is defined over the interval containing $x = 0$. In this case, the interval is $x < 1$, so we will use the sub-function $f(x) = x^2$.

To calculate the value of $f(0)$, we substitute $x = 0$ into the sub-function $f(x) = x^2$:

$$f(0) = (0)^2 = 0$$

Example 3

Determine the value of $f(3)$ for the piecewise function:

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

Solution

To determine the value of $f(3)$, we need to find the sub-function that is defined over the interval containing $x = 3$. In this case, the interval is $x \geq 3$, so we will use the sub-function $f(x) = 3x + 2$.

To calculate the value of $f(3)$, we substitute $x = 3$ into the sub-function $f(x) = 3x + 2$:

$$f(3) = 3(3) + 2 = 9 + 2 = 11$$

Final Thoughts

Q: What is a piecewise function?

A: A piecewise function is a function that is defined by multiple sub-functions, each applied to a specific interval of the domain. These sub-functions are often referred to as "pieces" of the function, and they are used to define the function's behavior over different intervals.

Q: How do I determine the value of a piecewise function at a specific point?

A: To determine the value of a piecewise function at a specific point, you need to find the sub-function that is defined over the interval containing the point, and then calculate the value of the function at that point.

Q: What if the point is not in any of the intervals?

A: If the point is not in any of the intervals, then the function is not defined at that point. In other words, the function has a "gap" or a "hole" at that point.

Q: Can I have multiple sub-functions with the same interval?

A: No, you cannot have multiple sub-functions with the same interval. Each interval must have a unique sub-function.

Q: How do I know which sub-function to use?

A: To determine which sub-function to use, you need to find the interval that contains the point. Once you have found the interval, you can use the corresponding sub-function to calculate the value of the function at that point.

Q: Can I have a piecewise function with an infinite number of sub-functions?

A: Yes, you can have a piecewise function with an infinite number of sub-functions. However, this is not a common occurrence in mathematics.

Q: How do I graph a piecewise function?

A: To graph a piecewise function, you need to graph each sub-function separately, and then combine the graphs to form the final graph of the piecewise function.

Q: Can I have a piecewise function with a constant sub-function?

A: Yes, you can have a piecewise function with a constant sub-function. For example, the function $f(x) = \begin{cases} 2, & x < 1 \\ 3, & x \geq 1 \end{cases}$ has a constant sub-function $f(x) = 2$ for $x < 1$.

Q: Can I have a piecewise function with a linear sub-function?

A: Yes, you can have a piecewise function with a linear sub-function. For example, the function $f(x) = \begin{cases} 2x, & x < 1 \\ 3x + 2, & x \geq 1 \end{cases}$ has a linear sub-function $f(x) = 2x$ for $x < 1$.

Q: Can I have a piecewise function with a quadratic sub-function?

A: Yes, you can have a piecewise function with a quadratic sub-function. For example, the function $f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1 \end{cases}$ has a quadratic sub-function $f(x) = x^2$ for $x < 1$.

Q: Can I have a piecewise function with a polynomial sub-function?

A: Yes, you can have a piecewise function with a polynomial sub-function. For example, the function $f(x) = \begin{cases} x^3, & x < 1 \\ 2x^2 + 1, & x \geq 1 \end{cases}$ has a polynomial sub-function $f(x) = x^3$ for $x < 1$.

Q: Can I have a piecewise function with a rational sub-function?

A: Yes, you can have a piecewise function with a rational sub-function. For example, the function $f(x) = \begin{cases} \frac{1}{x}, & x < 1 \\ \frac{2}{x} + 1, & x \geq 1 \end{cases}$ has a rational sub-function $f(x) = \frac{1}{x}$ for $x < 1$.

Q: Can I have a piecewise function with a trigonometric sub-function?

A: Yes, you can have a piecewise function with a trigonometric sub-function. For example, the function $f(x) = \begin{cases} \sin(x), & x < 1 \\ \cos(x) + 1, & x \geq 1 \end{cases}$ has a trigonometric sub-function $f(x) = \sin(x)$ for $x < 1$.

Q: Can I have a piecewise function with an exponential sub-function?

A: Yes, you can have a piecewise function with an exponential sub-function. For example, the function $f(x) = \begin{cases} e^x, & x < 1 \\ e^{2x} + 1, & x \geq 1 \end{cases}$ has an exponential sub-function $f(x) = e^x$ for $x < 1$.

Q: Can I have a piecewise function with a logarithmic sub-function?

A: Yes, you can have a piecewise function with a logarithmic sub-function. For example, the function $f(x) = \begin{cases} \log(x), & x < 1 \\ \log(2x) + 1, & x \geq 1 \end{cases}$ has a logarithmic sub-function $f(x) = \log(x)$ for $x < 1$.

Q: Can I have a piecewise function with a combination of different types of sub-functions?

A: Yes, you can have a piecewise function with a combination of different types of sub-functions. For example, the function $f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & 1 \leq x < 2 \\ \sin(x), & x \geq 2 \end{cases}$ has a combination of quadratic, linear, and trigonometric sub-functions.

Q: Can I have a piecewise function with an infinite number of sub-functions?

A: Yes, you can have a piecewise function with an infinite number of sub-functions. However, this is not a common occurrence in mathematics.

Q: Can I have a piecewise function with a sub-function that is not a function?

A: No, you cannot have a piecewise function with a sub-function that is not a function. Each sub-function must be a function.

Q: Can I have a piecewise function with a sub-function that is a constant function?

A: Yes, you can have a piecewise function with a sub-function that is a constant function. For example, the function $f(x) = \begin{cases} 2, & x < 1 \\ 3, & x \geq 1 \end{cases}$ has a constant sub-function $f(x) = 2$ for $x < 1$.

Q: Can I have a piecewise function with a sub-function that is a linear function?

A: Yes, you can have a piecewise function with a sub-function that is a linear function. For example, the function $f(x) = \begin{cases} 2x, & x < 1 \\ 3x + 2, & x \geq 1 \end{cases}$ has a linear sub-function $f(x) = 2x$ for $x < 1$.

Q: Can I have a piecewise function with a sub-function that is a quadratic function?

A: Yes, you can have a piecewise function with a sub-function that is a quadratic function. For example, the function $f(x) = \begin{cases} x^2, & x < 1 \\ 2x + 1, & x \geq 1 \end{cases}$ has a quadratic sub-function $f(x) = x^2$ for $x < 1$.

Q: Can I have a piecewise function with a sub-function that is a polynomial function?

A: Yes, you can have a piecewise function with a sub-function that is a polynomial function. For example, the function $f(x) = \begin{cases} x^3, & x < 1 \\ 2x^2 + 1, & x \geq 1 \end{cases}$ has a polynomial sub-function $f(x) = x^3$ for $x < 1$.

Q: Can I have a piecewise function with a sub-function that is a rational function?

A: Yes, you can have a piecewise function with a sub-function that is a rational function. For example, the function $f(x) = \begin{cases} \frac{1}{x}, & x < 1 \\ \frac{2}{x} + 1, & x \geq 1 \end{cases}$ has a rational sub-function $f(x) = \frac{1}{x}$ for $x < 1$.

**Q: Can I have a piecewise function with a sub-function that is