Problem 30 - Algebra 2 and Trigonometry

Question

In \(23-34,\) evaluate each function for the given value. Be sure to show your work. $$ \mathrm{f}(x)=\left(3 x^{-3}-2 x^{-3}\right)^{2} ; \mathrm{f}(-2) $$

Step-by-Step Solution

Answer

The function evaluated at \(-2\) is \( \frac{1}{64} \).

Step 1: Substitute the value into the function

Substitute \( x = -2 \) into the function \( f(x) = (3x^{-3} - 2x^{-3})^2 \). This gives us:\[ f(-2) = (3(-2)^{-3} - 2(-2)^{-3})^2 \]

Step 2: Evaluate the expression inside the parentheses

Simplify the expression inside the parentheses by calculating the exponents. Recall that \((-2)^{-3} = -\frac{1}{8}\). Therefore:\[ 3(-2)^{-3} = 3 \times -\frac{1}{8} = -\frac{3}{8} \]\[ 2(-2)^{-3} = 2 \times -\frac{1}{8} = -\frac{2}{8} = -\frac{1}{4} \]

Step 3: Simplify the expression

Subtract the two fractions inside the parentheses:\[ -\frac{3}{8} - (-\frac{1}{4}) = -\frac{3}{8} + \frac{1}{4} \]

Step 4: Combine fractions

Find a common denominator, which is 8. Thus, convert \(\frac{1}{4} = \frac{2}{8}\):\[ -\frac{3}{8} + \frac{2}{8} = -\frac{1}{8} \]

Step 5: Square the result

Square the result from Step 4:\[ \left(-\frac{1}{8}\right)^2 = \frac{1}{64} \]

Step 6: Provide the final answer

The function \( f(x) \) evaluated at \( x = -2 \) is \( \frac{1}{64} \). Therefore, \( f(-2) = \frac{1}{64} \).

Key Concepts

Function Evaluation

Function evaluation involves substituting a specific value for the variable in a function. It is like having a formula and plugging in a number to see what you get. Imagine a recipe where you add different quantities to get the same dish. In this exercise, our dish is the function \( f(x) = (3x^{-3} - 2x^{-3})^2 \), and the key ingredient is \( x = -2 \).

• Start by substituting \( x = -2 \) into the function.
• This means you replace every \( x \) in \( f(x) \) with \( -2 \).

Substitution helps us see how the output changes with different inputs. For our function, substituting \( -2 \) gives us the initial expression \( f(-2) = (3(-2)^{-3} - 2(-2)^{-3})^2 \). The key takeaway here is understanding that function evaluation is like testing how a rule or formula plays out when you give it certain numbers.

Exponents and Powers

Exponents can be thought of as instructions for repeated multiplication or, in the case of negative exponents, for division involving powers of numbers. In this problem, we deal with powers of \( -2 \).When we see \( (-2)^{-3} \), it is telling us to take the reciprocal of \((-2)^3\). So, \((-2)^{-3} = \frac{1}{(-2)^3} = -\frac{1}{8} \). The negative sign in the exponent signifies "flip over the fraction," and the number that the exponent is applied to is \(-2\).In mathematical practice,

• First, calculate the base raised to the positive power: \((-2)^3 = -8\).
• Then, take the reciprocal to get: \(-\frac{1}{8}\).

Understanding exponents is crucial when it comes to simplifying terms in polynomial functions, as they often signify the growth or rate of change. In polynomial equations, manipulating these can alter the function's behavior significantly.

Fraction Operations

Fractions, which represent parts of a whole, are key components in mathematics. Understanding how to manipulate fractions is essential, especially when simplifying expressions in functions.In our exercise, we end up with fractions during the evaluation process: \(-\frac{3}{8}\) and \(-\frac{1}{4}\). Performing operations on these fractions involves knowing how to find common denominators.

• The common denominator allows us to "speak the same language" for comparison. Here, the common denominator is 8.
• Convert \(-\frac{1}{4}\) to \(-\frac{2}{8}\) so both fractions have the denominator of 8.
• Add or subtract to simplify: \(-\frac{3}{8} + \frac{2}{8} = -\frac{1}{8}\).

Fraction operations are the backbone for many math problems, enabling us to maintain balance and consistency across equations. This knowledge is essential in ensuring calculations are accurate, as shown by our result which is then squared to get \(\frac{1}{64}\). Mastering these enables ease in dealing with complex polynomial function evaluations.