Problem 24 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

Evaluated values are: \( f(0)=0 \), \( f(1)=-3 \), \( f(-1)=-5 \), \( f\left(\frac{3}{2}\right)=-\frac{45}{8} \), \( f\left(\frac{x}{2}\right)=\frac{x^3}{8} - x^2 \), \( f(x^2)=x^6 - 4x^4 \).

Step 1: Evaluate f(0)

Substitute \( x = 0 \) into the function \( f(x) = x^3 - 4x^2 \). Compute:\[f(0) = 0^3 - 4(0)^2 = 0 - 0 = 0\]

Step 2: Evaluate f(1)

Substitute \( x = 1 \) into the function. Compute:\[f(1) = 1^3 - 4(1)^2 = 1 - 4 = -3\]

Step 3: Evaluate f(-1)

Substitute \( x = -1 \) into the function. Compute:\[f(-1) = (-1)^3 - 4(-1)^2 = -1 - 4 = -5\]

Step 4: Evaluate f(\(\frac{3}{2}\))

Substitute \( x = \frac{3}{2} \) into the function. Compute:\[f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^3 - 4\left(\frac{3}{2}\right)^2\]Calculate each term:\[\left(\frac{3}{2}\right)^3 = \frac{27}{8}, \quad \left(\frac{3}{2}\right)^2 = \frac{9}{4}, \quad 4\left(\frac{3}{2}\right)^2 = 4 \cdot \frac{9}{4} = 9\]Substituting back:\[f\left(\frac{3}{2}\right) = \frac{27}{8} - 9 = \frac{27}{8} - \frac{72}{8} = -\frac{45}{8}\]

Step 5: Evaluate f(\(\frac{x}{2}\))

Substitute \( x = \frac{x}{2} \) into the function. Compute:\[f\left(\frac{x}{2}\right) = \left(\frac{x}{2}\right)^3 - 4\left(\frac{x}{2}\right)^2\]Calculate each term:\[\left(\frac{x}{2}\right)^3 = \frac{x^3}{8}, \quad \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}, \quad 4\left(\frac{x}{2}\right)^2 = 4 \cdot \frac{x^2}{4} = x^2\]Substituting back:\[f\left(\frac{x}{2}\right) = \frac{x^3}{8} - x^2\]

Step 6: Evaluate f(\(x^2\))

Substitute \( x = x^2 \) into the function. Compute:\[f(x^2) = (x^2)^3 - 4(x^2)^2\]Calculate each term:\[(x^2)^3 = x^6, \quad 4(x^2)^2 = 4x^4\]Substituting back:\[f(x^2) = x^6 - 4x^4\]

Step 7: Conclusion

We have now evaluated the function \( f(x) = x^3 - 4x^2 \) at all the indicated values. The results of these computations are:- \( f(0) = 0 \)- \( f(1) = -3 \)- \( f(-1) = -5 \)- \( f\left(\frac{3}{2}\right) = -\frac{45}{8} \)- \( f\left(\frac{x}{2}\right) = \frac{x^3}{8} - x^2 \)- \( f(x^2) = x^6 - 4x^4 \)

Key Concepts

Polynomial Functions

Polynomial functions are expressions that include terms like constants and variables raised to non-negative integer exponents. The function \( f(x) = x^3 - 4x^2 \) is a typical example of a polynomial function. It contains:

• Three terms: \( x^3 \) (a cubic term) and \( -4x^2 \) (a quadratic term).
• The coefficients 1 for \( x^3 \) and -4 for \( x^2 \).

A polynomial can have one or more terms, and each term contains a variable raised to an exponent and a coefficient. In this case, the variable is \( x \), and the highest degree of the polynomial is the highest exponent, which is 3 in the cubic term. This determines the name and certain properties of the polynomial, such as its graph shape. Understanding polynomials helps us evaluate them at specific points, which is essential for various mathematical and real-life applications.

Substitution Method

The substitution method is a straightforward way to find the value of a function at given points. To use this method, substitute a specific value for the variable \( x \) in the function and then simplify the resulting expression. For instance, if you want to calculate \( f(1) \) for \( f(x) = x^3 - 4x^2 \), replace \( x \) with 1: \[ f(1) = 1^3 - 4(1)^2 = 1 - 4 = -3 \] By doing this step-by-step replacement and calculation for each value or expression, you get the function's specific output. The substitution method is a versatile tool used not only in mathematics but also in various scientific fields where modeling and predictions are necessary.

Exponents and Powers

Exponents and powers are a fundamental part of polynomial functions. Exponents tell us how many times to multiply a base number by itself. In the expression \( x^3 \), the base is \( x \), and the exponent is 3, meaning \( x \) is multiplied by itself three times: \( x \times x \times x \). Here's why understanding exponents is important:

• They allow us to express large numbers or repeated multiplication in a compact form.
• They determine the degree of polynomial terms, which affects the polynomial's behavior and graph.
• Simple rules for operations with exponents help simplify expressions: for instance, \( (x^a)^b = x^{a \times b} \).

In our function \( f(x) = x^3 - 4x^2 \), understanding how to work with the exponents is crucial for correctly evaluating the function at various points.

Function Notation

Function notation, such as \( f(x) \), is a way to represent functions in a clear and concise manner. It's a symbolic way to link an input with its corresponding output. In this format:

• \( f \) is the name of the function.
• \( x \) within parentheses is the input variable or argument of the function.
• The entire expression \( f(x) \) denotes the value of the function at \( x \).

When you see \( f(0), f(1), \) or \( f(a) \), these denote function values at specific points. It simplifies communication and calculation in mathematics, making it easier to work with different functional models. Understanding this notation is key to misinterpreting expressions and solutions when evaluating functions.