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) Approximately -3.29; (b) Approximately 1.20; (c) Undefined.

Step 1: Evaluate f(0.79)

Substitute the value 0.79 into the function \( f(x) = \frac{\sqrt{x^2 + 9}}{x - \sqrt{3}} \). This becomes \( f(0.79) = \frac{\sqrt{(0.79)^2 + 9}}{0.79 - \sqrt{3}} \). Calculate the expressions inside the square root and the denominator separately. \( (0.79)^2 = 0.6241 \), and \( \sqrt{0.6241 + 9} = \sqrt{9.6241} \). The denominator \( 0.79 - \sqrt{3} \) is approximately \( 0.79 - 1.732 \approx -0.942 \). After calculating, \( f(0.79) \approx \frac{3.102 \text{ (approx)}}{-0.942} \). Solve this to get \( f(0.79) \approx -3.29 \).

Step 2: Evaluate f(12.26)

Substitute the value 12.26 into the function as \( f(12.26) = \frac{\sqrt{(12.26)^2 + 9}}{12.26 - \sqrt{3}} \). First, calculate \( 12.26^2 = 150.3076 \), so \( \sqrt{150.3076 + 9} = \sqrt{159.3076} \). The denominator is \( 12.26 - \sqrt{3} \), approximately \( 12.26 - 1.732 \approx 10.528 \). Solving this gives \( f(12.26) \approx \frac{12.619 \text{ (approx)}}{10.528} \). After division, \( f(12.26) \approx 1.1999 \).

Step 3: Evaluate f(\sqrt{3})

Substitute \( x=\sqrt{3} \) into the function: \( f(\sqrt{3}) = \frac{\sqrt{(\sqrt{3})^2 + 9}}{\sqrt{3} - \sqrt{3}} \). Simplifying inside the square root, \( \sqrt{3}^2 = 3 \), so it becomes \( \frac{\sqrt{3 + 9}}{0} = \frac{\sqrt{12}}{0} \). Division by zero is undefined, therefore \( f(\sqrt{3}) \) is undefined.

Key Concepts

Functions Evaluation

Understanding how to evaluate functions is a cornerstone of calculus and algebra. A function, like the one in our exercise, describes a relationship between input and output. Here, the function is defined as \( f(x) = \frac{\sqrt{x^2 + 9}}{x-\sqrt{3}} \). The goal is to find the output of the function given specific input values such as 0.79, 12.26, and \( \sqrt{3} \).

To evaluate this function:

• Substitute the input value into the function wherever there is an \( x \).
• Carry out the arithmetic operations step by step, following the order of operations (parentheses, Exponents/roots, multiplication/division).
• Simplify the expression fully to reach the final value.

While this process might look simple on paper, it often requires attention to detail to ensure each calculation is exact. It is crucial to double-check each step, especially when exponents and roots are involved.

Division by Zero in Calculus

In calculus, division by zero is a situation we always need to avoid. Division by zero does not have a well-defined meaning mathematically, because you cannot divide a number into zero parts.

In our exercise, when evaluating \( f(\sqrt{3}) \), the denominator \( x - \sqrt{3} \) becomes zero. This happens because the expression resolves to \( \sqrt{3} - \sqrt{3} \), which equals zero.

Since division by zero is undefined, the function at \( f(\sqrt{3}) \) is also undefined.

This highlights the importance of careful calculations and understanding which input values can make the function undefined, or discontinuous. When encountering such values, they are usually either avoided or further analyzed, especially in more advanced calculus topics, for insights into the behavior of the function.

Square Root Calculations

Square roots often appear in functions, and they come with their own set of calculations and challenges. The square root of a number is a value that, when multiplied by itself, gives the original number.

In our function, \( f(x) = \frac{\sqrt{x^2 + 9}}{x-\sqrt{3}} \), the square root is applied to \( x^2 + 9 \). To evaluate:

• Square the number first (as in \( x^2 \)), then add any other numbers inside the root.
• Calculate the square root of the resulting sum.

For example, when evaluating \( f(0.79) \), after calculating \( 0.79^2 + 9 \), the square root is taken of the resulting 9.6241, yielding approximately 3.102. This process reflects the step-by-step nature of dealing with square roots and ensuring all arithmetic is precise.

Mistakes often happen while taking square roots, so it's advised to use a calculator for accuracy. Understanding these calculations is essential for tackling not only basic algebra problems but also more complex calculus problems involving derivatives and integrals where square roots appear frequently.