Problem 68 - Elementary and Intermediate Algebra

Question

Use a calculator to find each function value. Round to the nearest ten- thousandth. See Example 5 and Using Your Calculator \(g(x)=\sqrt{7-4 x}\) a. \(g(-\pi)\) b. \(g(0.5)\)

Step-by-Step Solution

Answer

a. \(g(-\pi) \approx 4.4221\); b. \(g(0.5) \approx 2.2361\).

Step 1: Evaluate the function for \( g(-\pi) \)

First, substitute \(-\pi\) for \(x\) in the function \(g(x)=\sqrt{7-4x}\). This gives us: \[ g(-\pi) = \sqrt{7 - 4(-\pi)} = \sqrt{7 + 4\pi} \]. Calculate this using a calculator.

Step 2: Calculate \( \sqrt{7 + 4\pi} \)

Using your calculator, find \( \pi \) as approximately 3.1416. Then substitute:\[ 4 \times \pi \approx 4 \times 3.1416 = 12.5664 \].Now, add 7 to 12.5664 to get 19.5664. Finally, find the square root: \( \sqrt{19.5664} \approx 4.4221 \).Thus, \( g(-\pi) \approx 4.4221 \).

Step 3: Evaluate the function for \( g(0.5) \)

Substitute \(0.5\) for \(x\) in the function \(g(x)=\sqrt{7-4x}\). This gives us: \[ g(0.5) = \sqrt{7 - 4(0.5)} = \sqrt{7 - 2} \]. Calculate this using a calculator.

Step 4: Calculate \( \sqrt{7 - 2} \)

Compute the inside of the square root first: \[ 7 - 2 = 5 \].Now find the square root using a calculator: \( \sqrt{5} \approx 2.2361 \).Therefore, \( g(0.5) \approx 2.2361 \).

Key Concepts

Square Root Calculation

The square root of a number is a value that, when multiplied by itself, gives the number. For instance, the square root of 9 is 3, because 3 times 3 equals 9. The square root is denoted by the symbol \( \sqrt{} \). Finding the square root of a number is fundamental in mathematics and appears in various problems and equations. Br<>To calculate the square root in our example, we have the expression \( \sqrt{7 - 4x} \). First, substitute the value for \( x \), perform the arithmetic operations inside the square root, and then compute the square root of that result. In simpler exercises, such as finding \( \sqrt{16} \), you might know the result by heart. However, in more complex scenarios like \( \sqrt{19.5664} \), using a calculator becomes necessary.
Square root calculations are pervasive in algebra and have practical applications in geometry, physics, and engineering, helping to solve for dimensions, analyze forces, and more.

Using a Calculator

Calculators are handy tools, especially when dealing with complex mathematical operations like square roots, exponentiations, and logarithms. When using a calculator to find square roots, ensure you understand its order of operations to achieve the accurate result.
When finding \( g(-\pi) \), it involves inputting \( 7 + 4\times\pi \) correctly. Always perform multiplication first, then addition, before taking the square root. In our calculation, remember to:

• First, compute \( 4\times\pi \), using \( \pi \) approximately 3.1416.
• Add the result to 7 to get the total sum inside the square root.
• Finally, use the square root function to find the correct value.

Calculators simplify evaluations when numbers are too large or cumbersome to deal with mentally, ensuring accuracy in your mathematical work.

Rounding Numbers

Rounding is crucial in mathematics to present figures in a more manageable or readable form without much loss of precision. The concept of rounding involves simplifying a number by keeping its significant figures and omitting less significant ones.
To round numbers to the nearest ten-thousandth, as in this exercise, you look at the fifth decimal place. If it's 5 or greater, round up the fourth decimal place by one. If it's less than 5, keep it unchanged. For instance, to round 4.422139 to the nearest ten-thousandth, observe the 5th decimal place, 3. Since it's less than 5, the rounded number is 4.4221. Similarly, for 2.236067, the 5th place is 6, prompting rounding up of the fourth decimal place, yielding 2.2361.
Rounding helps in approximating values for easier communication and fewer errors in calculations, especially in scientific, financial, and statistical data where precision is pivotal.