/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Evaluate \(f(x)\) at the given \... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 49

Evaluate \(f(x)\) at the given \(x\). Approximate each result to the nearest hundredth. $$f(x)=x^{3 / 2}-x^{1 / 2}, \quad x=50$$

Short Answer

Expert verified
\(f(50) \approx 346.48\).

Step by step solution

01

Substitute the Value of x

To evaluate the function \(f(x) = x^{3/2} - x^{1/2}\) at \(x=50\), start by substituting 50 for \(x\) in the expression.
02

Evaluate x to the Power of 3/2

Calculate \(50^{3/2}\). The power \(3/2\) can be interpreted as the square root and then the cube, so calculate \(\sqrt{50}\) first.
03

Calculate Square Root of 50

Find the square root of 50, \(\sqrt{50} \approx 7.071\).
04

Cube the Result of the Square Root

Raise the square root result to the power of 3: \((7.071)^3 \approx 353.55\).
05

Evaluate x to the Power of 1/2

Calculate \(50^{1/2}\) which is simply the square root of 50, hence: \(\sqrt{50} \approx 7.071\).
06

Subtract the Values

Subtract the value from Step 5 \(\sqrt{50} \approx 7.071\) from the value in Step 4 \((7.071)^3 \approx 353.55\): \[f(50) = 353.55 - 7.071 \approx 346.48\]
07

Round to the Nearest Hundredth

Round the result from Step 6 to the nearest hundredth: \[f(50) \approx 346.48\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponentiation
Exponentiation is a mathematical operation that involves raising a number to a power. In this exercise, you encountered the expression \( x^{3/2} \), which can be a little tricky to understand without breaking it down. Here's a simple explanation:
  • The base, in this case 50, is the number that is being multiplied by itself.
  • The exponent 3/2 is a fraction, meaning we first find the root of the base and then raise the result to an integer power.
  • To solve \( 50^{3/2} \), we can think of it as \( (50^{1/2})^3 \), which means we first find the square root of 50 and then cube that result.
Understanding fractional exponents is crucial as it allows us to solve more complex expressions. Practicing with different bases and exponents will help strengthen your foundational knowledge.
Square Root
The square root of a number is one of its two equal factors. When you see \( x^{1/2} \), you're being asked to find the square root of \( x \). In our exercise, this was part of the process:
  • We had to calculate \( \sqrt{50} \) to evaluate \( 50^{1/2} \) and also to work towards \( 50^{3/2} \).
  • Square roots come up often in areas like geometry, physics, and statistics, making it a valuable concept to understand.
  • Numerically approximating square roots like \( \sqrt{50} \approx 7.071 \) is common in real-world applications where exact calculations are impractical.
To get better at estimating square roots, remember that if a number is between two perfect squares, its square root will lie between the square roots of those perfect squares.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations. The function given in the exercise, \( f(x) = x^{3/2} - x^{1/2} \), is an algebraic expression because it includes exponents, a variable \( x \), and subtraction.
  • First, substitute the specific value for the variable. In our exercise, we plugged \( x = 50 \) into the expression.
  • Next, each part of the expression is evaluated separately before any addition or subtraction. This means finding \( x^{3/2} \) and \( x^{1/2} \) individually for \( x=50 \).
  • Finally, solve the expression by performing the subtraction: \( 353.55 - 7.071 = 346.48 \), which was the function evaluation at \( x = 50 \).
Understanding how to manipulate algebraic expressions and substitute variables is key when solving a wide range of mathematical problems. With ongoing practice, it becomes much easier to manage these calculations.

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