Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

For \(f(x)=\sqrt{x^{2}+9} /(x-\sqrt{3})\), find each value. (a) \(f(0.79)\) (b) \(f(12.26)\) (c) \(f(\sqrt{3})\)

Step-by-Step Solution

Answer

(a) \( f(0.79) \approx -3.292 \), (b) \( f(12.26) \approx 1.198 \), (c) \( f(\sqrt{3}) \) is undefined.

Step 1: Evaluate expression inside the square root

First, evaluate the expression inside the square root for each function input. This means calculating \( x^2 + 9 \) for the given values of \( x \).

Step 2: Substitute 0.79 into the function

Calculate \( x^2 + 9 \) for \( x = 0.79 \): \((0.79)^2 + 9 = 0.6241 + 9 = 9.6241.\)

Step 3: Calculate the square root for 0.79

Take the square root of the result obtained: \(\sqrt{9.6241} \approx 3.1023.\)

Step 4: Calculate denominator and function for 0.79

Calculate the denominator with \( x = 0.79 \): \(0.79 - \sqrt{3} \approx 0.79 - 1.732 = -0.942.\) Then, compute the function: \( f(0.79) = \frac{3.1023}{-0.942} \approx -3.292. \)

Step 5: Substitute 12.26 into the function

Calculate \( x^2 + 9 \) for \( x = 12.26 \): \((12.26)^2 + 9 = 150.3076 + 9 = 159.3076.\)

Step 6: Calculate the square root for 12.26

Take the square root of the result obtained: \( \sqrt{159.3076} \approx 12.6193. \)

Step 7: Calculate denominator and function for 12.26

Calculate the denominator with \( x = 12.26 \): \(12.26 - \sqrt{3} \approx 12.26 - 1.732 = 10.528.\) Then, compute the function: \( f(12.26) = \frac{12.6193}{10.528} \approx 1.198. \)

Step 8: Evaluate the function for \( x=\sqrt{3} \)

Since \( x = \sqrt{3} \) causes the denominator \( x - \sqrt{3} \) to equal zero, \( f(\sqrt{3}) \) is undefined.

Key Concepts

Function Evaluation

In calculus, function evaluation is an essential concept. It involves calculating the value of a function at a specific point, often denoted as \( f(x) \). To evaluate \( f(x) \), you substitute the given value of \( x \) into the function's equation. For example, with \( f(x) = \frac{\sqrt{x^2 + 9}}{(x - \sqrt{3})} \), evaluating \( f(0.79) \) means replacing \( x \) with \( 0.79 \) throughout the entire function expression.
We work step-by-step:

• Substitute the number into the function to replace the variable \( x \).
• Perform the arithmetic operations inside any sub-expressions, like squares or additions.
• Simplify the expression as much as possible before moving on to more complex operations.

Function evaluation is fundamental for understanding how a function behaves at different points and determining specific output values.

Square Root Calculation

Square root calculation is another crucial aspect of mathematics, particularly in solving calculus problems. Taking a square root reverses the operation of squaring a number. In our function, \( \sqrt{x^2 + 9} \), we need to simplify \( x^2 + 9 \) first before taking the square root.
Here's how you approach it:

• Calculate \( x^2 \) to get the value of squared \( x \).
• Add the constant (in this case, 9) to \( x^2 \).
• Apply the square root to the resulting sum.

For example, if \( x = 12.26 \), we find \((12.26)^2 + 9 \) and then take the square root of that result. Remember that the square root function produces a positive result for real numbers.

Undefined Expressions

An expression becomes undefined when a mathematical operation within the expression cannot be completed in the real number system. Most commonly, division by zero causes an expression to be undefined. In the function \( f(x) = \frac{\sqrt{x^2 + 9}}{(x - \sqrt{3})} \), substituting \( x = \sqrt{3} \) leads to the denominator being zero, making the function undefined at that point.
Understanding undefined expressions is critical:

• Avoid dividing by zero; monitor equations to ensure no zero denominator occurs.
• Identify and label values that make the denominator zero as points of discontinuity or non-existence in the function.
• Be aware that undefined expressions restrict the possible input values for a function.

Knowing where a function is undefined helps in graphing and analyzing the function's behavior across its domain.