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Chapter 2: Problem 18

Evaluate the function at the indicated values. $$\begin{aligned}&f(x)=x^{3}-4 x^{2};\\\&f(0), f(1), f(-1), f\left(\frac{3}{2}\right), f\left(\frac{x}{2}\right), f\left(x^{2}\right)\end{aligned}$$

Short Answer

Expert verified
\( f(0) = 0 \); \( f(1) = -3 \); \( f(-1) = -5 \); \( f(\frac{3}{2}) = -\frac{45}{8} \); \( f(\frac{x}{2}) = \frac{x^3}{8} - x^2 \); \( f(x^2) = x^6 - 4x^4 \).

Step by step solution

01

Understanding the Given Function

The function provided is \( f(x) = x^3 - 4x^2 \). We need to evaluate this function at specified values of \( x \): 0, 1, -1, \( \frac{3}{2} \), \( \frac{x}{2} \), and \( x^2 \).
02

Evaluating at x = 0

Substitute \( x = 0 \) into the function: \( f(0) = 0^3 - 4 \times 0^2 = 0 - 0 = 0 \). Thus, \( f(0) = 0 \).
03

Evaluating at x = 1

Substitute \( x = 1 \) into the function: \( f(1) = 1^3 - 4 \times 1^2 = 1 - 4 = -3 \). Thus, \( f(1) = -3 \).
04

Evaluating at x = -1

Substitute \( x = -1 \) into the function: \( f(-1) = (-1)^3 - 4 \times (-1)^2 = -1 - 4 = -5 \). Thus, \( f(-1) = -5 \).
05

Evaluating at x = \( \frac{3}{2} \)

Substitute \( x = \frac{3}{2} \) into the function: \( f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^3 - 4 \times \left(\frac{3}{2}\right)^2 = \frac{27}{8} - 4 \times \frac{9}{4} = \frac{27}{8} - 9 = \frac{27}{8} - \frac{72}{8} = -\frac{45}{8} \). Thus, \( f\left(\frac{3}{2}\right) = -\frac{45}{8} \).
06

Evaluating at x = \( \frac{x}{2} \)

Substitute \( x = \frac{x}{2} \) into the function: \( f\left(\frac{x}{2}\right) = \left(\frac{x}{2}\right)^3 - 4 \times \left(\frac{x}{2}\right)^2 = \frac{x^3}{8} - 4 \times \frac{x^2}{4} = \frac{x^3}{8} - x^2 \). Thus, \( f\left(\frac{x}{2}\right) = \frac{x^3}{8} - x^2 \).
07

Evaluating at x = \( x^2 \)

Substitute \( x = x^2 \) into the function: \( f(x^2) = (x^2)^3 - 4 \times (x^2)^2 = x^6 - 4x^4 \). Thus, \( f(x^2) = x^6 - 4x^4 \).

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Key Concepts

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

Polynomial Functions
Polynomial functions are a fundamental building block in precalculus. They are expressions made up of variables and coefficients, structured in terms involving sums of powers. For example, in the function \( f(x) = x^3 - 4x^2 \), the terms \( x^3 \) and \( -4x^2 \) are combined by addition or subtraction. This is characteristic of polynomial expressions. The highest power of the variable determines the degree of the polynomial, which is vital for understanding its behavior. Since our example has \( x^3 \) as the highest power, it is a cubic polynomial. These types of functions are critical for understanding more complex equations and appear frequently in various branches of mathematics.
Substitution Method
The substitution method is a simple technique used to evaluate algebraic expressions, like polynomial functions, at given values. This involves replacing the variable with a specific number or another expression. In the example function \( f(x) = x^3 - 4x^2 \), substituting different values for \( x \) allows you to determine the function's output at those points.
  • For \( x = 0 \), substitute 0 in for \( x \): \( 0^3 - 4 \cdot 0^2 = 0 \).
  • For \( x = 1 \), substitute 1 in for \( x \): \( 1^3 - 4 \cdot 1^2 = -3 \).
  • For \( x = -1 \), substitute -1 in for \( x \): \( (-1)^3 - 4 \cdot (-1)^2 = -5 \).
This method is universally applicable, making it a powerful tool for working with polynomial expressions in precalculus.
Precalculus
Precalculus serves as the foundation for understanding calculus, and it encompasses essential topics like polynomial functions and other higher-degree expressions. A key aspect of precalculus is making sense of how these mathematical concepts fit together before moving on to more complex calculus ideas. When working with functions, students learn to evaluate them, analyze graphs, and comprehend the broader applications of these equations. Exercises similar to evaluating \( f(x) = x^3 - 4x^2 \) at multiple points help students synthesize these principles. This foundational knowledge is crucial, as many later mathematical concepts rely on a solid grasp of precalculus fundamentals.
Math Problem Solving
Math problem solving is a skill enhanced significantly by exercises like function evaluation. The process requires a structured approach: understanding the problem, deciding on strategies such as substitution, and then applying these to find a solution.
  • Identify the function: Recognize what kind of function you are working with (e.g., polynomial).
  • Analyze the requirements: Know what values you need to substitute into the function.
  • Solve systematically: Substitute and calculate step by step.
These steps train the mind to tackle more intricate problems methodically. By practicing such methods in precalculus, students bolster their analytical skills, prepare for advanced topics, and gain confidence in their mathematical abilities.

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Most popular questions from this chapter

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