Problem 52 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. If \(f(x)=\sqrt{x},\) find \(\frac{f(x)-f(4)}{x-4},\) for \(x=5,4.5,\) and 4.1 Round results to three decimal places.

Step-by-Step Solution

Answer

For \( x = 5 \), result is 0.236. For \( x = 4.5 \), result is 0.121. For \( x = 4.1 \), result is 0.025.

Step 1: Define the function

Given the function is \( f(x) = \sqrt{x} \).

Step 2: Determine \( f(4) \)

Calculate \( f(4) = \sqrt{4} = 2 \).

Step 3: Simplify the expression

Rewrite the expression \( \frac{f(x) - f(4)}{x - 4} \) as \( \frac{\sqrt{x} - 2}{x - 4} \).

Step 4: Compute for \( x = 5 \)

Substitute \( x = 5 \) into the simplified expression: \( \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \). Calculate \( \sqrt{5} \approx 2.236 \). Thus, \( 2.236 - 2 = 0.236 \). Rounded to three decimal places, the result is 0.236.

Step 5: Compute for \( x = 4.5 \)

Substitute \( x = 4.5 \) into the simplified expression: \( \frac{\sqrt{4.5} - 2}{4.5 - 4} = \sqrt{4.5} - 2 \). Calculate \( \sqrt{4.5} \approx 2.121 \). Thus, \( 2.121 - 2 = 0.121 \). Rounded to three decimal places, the result is 0.121.

Step 6: Compute for \( x = 4.1 \)

Substitute \( x = 4.1 \) into the simplified expression: \( \frac{\sqrt{4.1} - 2}{4.1 - 4} = \sqrt{4.1} - 2 \). Calculate \( \sqrt{4.1} \approx 2.025 \). Thus, \( 2.025 - 2 = 0.025 \). Rounded to three decimal places, the result is 0.025.

Key Concepts

Function Evaluation

Function evaluation is an important concept in calculus and algebra. It means finding the output of a function for a specific input.
For instance, if we have a function defined as \( f(x) = \sqrt{x} \), and we want to evaluate this function at \( x = 4 \), we substitute 4 into the function. This becomes \( f(4) = \sqrt{4} \).

• The process involves substituting the value of \( x \) given in the problem directly into the function.
• Evaluate according to the arithmetic needed.

In our example, \( f(4) = \sqrt{4} = 2\). This step is crucial as it sets the basis for further calculations.

Simplifying Expressions

Simplifying expressions helps in making complex problems more manageable. When dealing with the difference quotient, start by rewriting the expression properly.
In the given exercise, we are asked to simplify \( \frac{f(x) - f(4)}{x - 4} \). First, recognize that \( f(x) \) is \( \sqrt{x} \) and \( f(4) = 2 \). Thus, it can be rewritten as:
\[ \frac{\sqrt{x} - 2}{x - 4} \]

• Identify the function values you need to subtract.
• Place them in a fraction with the correct denominator.

This simplified form makes it easier to deal with when substituting specific values for \( x \).

Square Roots

Understanding square roots is key to solving many algebra problems. A square root of a number \( y \) is a value that, when multiplied by itself, gives \( y \).
For example, \( \sqrt{4} = 2 \) because \( 2 * 2 = 4 \).

• Square roots always return non-negative values (the principal square root).
• When evaluating square roots, calculators are often used for non-perfect squares.

In our problem, we need to find square roots for various \( x \) values:

• For \( x = 5 \), \( \sqrt{5} \approx 2.236 \)
• For \( x = 4.5 \), \( \sqrt{4.5} \approx 2.121 \)
• For \( x = 4.1 \), \( \sqrt{4.1} \approx 2.025 \)

These calculations allow us to find the difference quotient for each \( x \).