Problem 16 - Contemporary Precalculus

Question

Find the average rate of change of the function f over the given interval. $$f(x)=\sqrt{x^{3}+2 x^{2}-6 x+5} \text { from } x=1 \text { to } x=1.00001$$

Step-by-Step Solution

Answer

Answer: The approximate average rate of change of the function f(x) over the given interval is 0.000024999.

Step 1: 1. Evaluate the function at the given points

First, we need to evaluate f(x) at x=1 and x=1.00001 For x=1, we have: $$f(1)=\sqrt{1^3+2(1)^2-6(1)+5}=\sqrt{1+2-6+5}=2$$ Now, for x=1.00001, we have: $$f(1.00001)=\sqrt{(1.00001)^3+2(1.00001)^2-6(1.00001)+5}≈2.000005$$

Step 2: 2. Apply the difference quotient formula

Now using the difference quotient formula, we will plug in our values of f(1) and f(1.00001) and solve for the average rate of change. $$\frac{f(b)-f(a)}{b-a}=\frac{f(1.00001)-f(1)}{1.00001-1}$$ = $$\frac{2.000005-2}{1.00001-1}=0.000024999$$ So, the average rate of change of the function f(x) over the given interval is approximately 0.000024999.

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus that is used to determine the average rate of change of a function over a specified interval. It's an important tool for understanding how functions behave as their input values change. This concept can be thought of as the slope of the secant line that connects two points on a graph of a function.

In mathematical terms, the difference quotient is expressed as:

• \( \frac{f(b) - f(a)}{b - a} \)

Here, \( f(a) \) and \( f(b) \) are the values of the function \( f \) at two different points \( a \) and \( b \).

• This formula gives us the average rate of change of the function \( f \) over the interval from \( a \) to \( b \).

It's a stepping stone to more advanced concepts such as the derivative, which measures the instantaneous rate of change.

Function Evaluation

Function evaluation is the process of determining the output of a function for a particular input. It involves substituting a specific value or variable into the function and solving it.

This step is crucial because it enables us to find exact values that are needed to apply the difference quotient.

• For example, given a function such as \( f(x) = \sqrt{x^3 + 2x^2 - 6x + 5} \), we can evaluate it at different points like \( x = 1 \) or \( x = 1.00001 \).
• By substituting these values into the function, we find that \( f(1) = 2 \) and an approximate value of \( f(1.00001) \approx 2.000005 \).

Function evaluation allows us to use these results to find other properties of the function, like its average rate of change.

Precalculus

Precalculus is a branch of mathematics that prepares students for calculus by covering essential concepts and tools. It often includes the study of functions, their properties, and manipulations.

Mastering precalculus is key to understanding more complex mathematical principles such as derivatives and integrals found in calculus.

• It helps students grasp the behavior and manipulation of different types of functions including algebraic, exponential, logarithmic, and trigonometric functions.
• Topics such as the difference quotient and function evaluation are foundational concepts covered in precalculus that pave the way to understanding calculus.

Through practice, students learn to solve a wide array of problems, thus gaining a stronger foundation for further mathematical studies.

Square Root Function

The square root function, typically represented as \( f(x) = \sqrt{x} \), is an important type of function where the output is the square root of the input value.

• For example, if \( x = 4 \), then \( f(x) = \sqrt{4} = 2 \).
• This function is defined only for non-negative values of \( x \) since square roots of negative numbers are not real numbers.

In the given exercise, the function \( f(x) = \sqrt{x^3 + 2x^2 - 6x + 5} \) incorporates more complex algebraic terms under the square root, leading to a more interesting behavior.

• The square root affects both the domain and the range of the function, meaning it impacts which \( x \) values can be used and what values \( f(x) \) can take on.

Understanding square root functions is important in precalculus to analyze and solve a variety of mathematical problems.