The Synthetic Substitution Algorithm

Oct 13, 2025 by wbrum.org 33 views

Introduction

Synthetic substitution is a technique used to evaluate functions at specific values of the variable. It is a powerful tool in algebra and calculus, allowing us to quickly and easily find the value of a function at a given point. In this article, we will explore how to use synthetic substitution to evaluate the function $f(x) = -x^3 + x^2 + 4x + 9$ at $x = -3$ .

The Synthetic Substitution Algorithm

The synthetic substitution algorithm is a step-by-step process that involves filling in a table with the values of the function and its derivatives. The table is set up as follows:

$x$	$f(x)$	$f'(x)$	$f''(x)$	$f'''(x)$
$-3$

To fill in the table, we need to evaluate the function and its derivatives at $x = -3$ . We will start by evaluating the function itself.

Evaluating the Function

To evaluate the function $f(x) = -x^3 + x^2 + 4x + 9$ at $x = -3$ , we substitute $x = -3$ into the function:

$f(-3) = -(-3)^3 + (-3)^2 + 4(-3) + 9$

Using the order of operations, we can simplify this expression:

$f(-3) = -(-27) + 9 - 12 + 9$

$f(-3) = 27 + 9 - 12 + 9$

$f(-3) = 33$

So, the value of the function at $x = -3$ is $f(-3) = 33$ .

Evaluating the Derivatives

To fill in the rest of the table, we need to evaluate the derivatives of the function at $x = -3$ . We will start by finding the first derivative of the function.

First Derivative

To find the first derivative of the function $f(x) = -x^3 + x^2 + 4x + 9$ , we will use the power rule of differentiation. The power rule states that if $f(x) = x^n$ , then $f'(x) = nx^{n-1}$ .

Using the power rule, we can find the first derivative of the function:

$f'(x) = -3x^2 + 2x + 4$

Now, we can evaluate the first derivative at $x = -3$ :

$f'(-3) = -3(-3)^2 + 2(-3) + 4$

$f'(-3) = -27 - 6 + 4$

$f'(-3) = -29$

Second Derivative

To find the second derivative of the function, we will differentiate the first derivative:

$f''(x) = -6x + 2$

Now, we can evaluate the second derivative at $x = -3$ :

$f''(-3) = -6(-3) + 2$

$f''(-3) = 18 + 2$

$f''(-3) = 20$

Third Derivative

To find the third derivative of the function, we will differentiate the second derivative:

$f'''(x) = -6$

The third derivative is a constant, so it is the same at all values of $x$ . Therefore, we can evaluate the third derivative at $x = -3$ :

$f'''(-3) = -6$

Filling in the Table

Now that we have evaluated the function and its derivatives at $x = -3$ , we can fill in the table:

$x$	$f(x)$	$f'(x)$	$f''(x)$	$f'''(x)$
$-3$	33	-29	20	-6

Conclusion

In this article, we used synthetic substitution to evaluate the function $f(x) = -x^3 + x^2 + 4x + 9$ at $x = -3$ . We filled in a table with the values of the function and its derivatives, and used the table to find the value of the function at $x = -3$ . The value of the function at $x = -3$ is $f(-3) = 33$ .

Discussion

Synthetic substitution is a powerful tool in algebra and calculus, allowing us to quickly and easily find the value of a function at a given point. It is a useful technique to have in your toolkit, and can be used to solve a wide range of problems. In this article, we used synthetic substitution to evaluate a function at a specific value of the variable, but it can also be used to find the value of a function at a critical point or to solve a system of equations.

Example Problems

Here are a few example problems that you can try using synthetic substitution:

Evaluate the function $f(x) = 2x^3 - 3x^2 + x + 1$ at $x = 2$ .
Evaluate the function $f(x) = x^4 - 2x^3 + 3x^2 - x + 1$ at $x = 1$ .
Evaluate the function $f(x) = x^5 - 4x^4 + 5x^3 - 6x^2 + 7x - 1$ at $x = 0$ .

I hope this helps! Let me know if you have any questions or need further clarification.

Introduction

Synthetic substitution is a powerful technique used to evaluate functions at specific values of the variable. In our previous article, we explored how to use synthetic substitution to evaluate the function $f(x) = -x^3 + x^2 + 4x + 9$ at $x = -3$ . In this article, we will answer some frequently asked questions about synthetic substitution.

Q: What is synthetic substitution?

A: Synthetic substitution is a technique used to evaluate functions at specific values of the variable. It involves filling in a table with the values of the function and its derivatives, and using the table to find the value of the function at the given point.

Q: How do I use synthetic substitution?

A: To use synthetic substitution, you need to follow these steps:

Evaluate the function at the given point.
Find the first derivative of the function.
Evaluate the first derivative at the given point.
Find the second derivative of the function.
Evaluate the second derivative at the given point.
Find the third derivative of the function.
Evaluate the third derivative at the given point.
Fill in the table with the values of the function and its derivatives.

Q: What is the purpose of synthetic substitution?

A: The purpose of synthetic substitution is to quickly and easily find the value of a function at a given point. It is a useful technique to have in your toolkit, and can be used to solve a wide range of problems.

Q: Can I use synthetic substitution to find the value of a function at a critical point?

A: Yes, you can use synthetic substitution to find the value of a function at a critical point. A critical point is a point where the function has a maximum or minimum value. To find the value of a function at a critical point, you need to follow the same steps as before, but you will need to use the second derivative to determine whether the point is a maximum or minimum.

Q: Can I use synthetic substitution to solve a system of equations?

A: Yes, you can use synthetic substitution to solve a system of equations. A system of equations is a set of equations that involve multiple variables. To solve a system of equations using synthetic substitution, you need to follow the same steps as before, but you will need to use the function and its derivatives to eliminate variables and solve for the remaining variables.

Q: What are some common mistakes to avoid when using synthetic substitution?

A: Some common mistakes to avoid when using synthetic substitution include:

Not evaluating the function at the given point.
Not finding the first derivative of the function.
Not evaluating the first derivative at the given point.
Not finding the second derivative of the function.
Not evaluating the second derivative at the given point.
Not finding the third derivative of the function.
Not evaluating the third derivative at the given point.

Q: How do I know if I have made a mistake when using synthetic substitution?

A: If you have made a mistake when using synthetic substitution, you may get an incorrect answer. To check your work, you can plug the values back into the original function and see if they are true. If they are not true, then you have made a mistake.

Q: Can I use synthetic substitution to evaluate functions with multiple variables?

A: Yes, you can use synthetic substitution to evaluate functions with multiple variables. However, you will need to use a different technique, such as partial derivatives, to find the derivatives of the function.

Q: Can I use synthetic substitution to evaluate functions with complex numbers?

A: Yes, you can use synthetic substitution to evaluate functions with complex numbers. However, you will need to use a different technique, such as complex differentiation, to find the derivatives of the function.

Conclusion

Synthetic substitution is a powerful technique used to evaluate functions at specific values of the variable. In this article, we answered some frequently asked questions about synthetic substitution. We hope this helps you to better understand how to use synthetic substitution to evaluate functions and solve problems.

Example Problems

Here are a few example problems that you can try using synthetic substitution:

Evaluate the function $f(x) = 2x^3 - 3x^2 + x + 1$ at $x = 2$ .
Evaluate the function $f(x) = x^4 - 2x^3 + 3x^2 - x + 1$ at $x = 1$ .
Evaluate the function $f(x) = x^5 - 4x^4 + 5x^3 - 6x^2 + 7x - 1$ at $x = 0$ .

I hope this helps! Let me know if you have any questions or need further clarification.