Problem 22 - Precalculus

Question

Use the given function \(f\) to find and simplify the following: \- \(f(2)\) \- \(2 f(a)\) \- \(f\left(\frac{2}{a}\right)\) \- \(f(-2)\) \- \(f(a+2)\) -\(\frac{f(a)}{2}\) \- \(f(2 a)\) \- \(f(a)+f(2)\) \- \(f(a+h)\) $$f(x)=3 x^{2}+3 x-2$$

Step-by-Step Solution

Answer

\(f(2)=16\), \(2f(a)=6a^2+6a-4\), \(f\left(\frac{2}{a}\right)=\frac{12}{a^2}+\frac{6}{a}-2\), \(f(-2)=4\), \(f(a+2)=3a^2+15a+16\), \(\frac{f(a)}{2}=\frac{3a^2}{2}+\frac{3a}{2}-1\), \(f(2a)=12a^2+6a-2\), \(f(a)+f(2)=3a^2+3a+14\), \(f(a+h)=3a^2+6ah+3h^2+3a+3h-2\).

Step 1: Evaluate \(f(2)\)

To find \(f(2)\), substitute \(x = 2\) into the function: \(f(x) = 3x^2 + 3x - 2\). Thus, \(f(2) = 3(2)^2 + 3(2) - 2 = 3 \times 4 + 6 - 2 = 12 + 6 - 2 = 16\).

Step 2: Compute \(2f(a)\)

Substitute \(x = a\) into the function to find \(f(a)\), then multiply the result by 2. \(f(a) = 3a^2 + 3a - 2\). So, \(2f(a) = 2(3a^2 + 3a - 2) = 6a^2 + 6a - 4\).

Step 3: Evaluate \(f\left(\frac{2}{a}\right)\)

Substitute \(x = \frac{2}{a}\) into the function. \(f\left(\frac{2}{a}\right) = 3\left(\frac{2}{a}\right)^2 + 3\left(\frac{2}{a}\right) - 2 = 3\left(\frac{4}{a^2}\right) + \frac{6}{a} - 2 = \frac{12}{a^2} + \frac{6}{a} - 2\).

Step 4: Evaluate \(f(-2)\)

Substitute \(x = -2\) into the function: \(f(x) = 3x^2 + 3x - 2\). Therefore, \(f(-2) = 3(-2)^2 + 3(-2) - 2 = 3 \times 4 - 6 - 2 = 12 - 6 - 2 = 4\).

Step 5: Evaluate \(f(a+2)\)

Substitute \(x = a+2\) into the function. \(f(a+2) = 3(a+2)^2 + 3(a+2) - 2\). First, expand \((a+2)^2 = a^2 + 4a + 4\). Then, \(3(a^2 + 4a + 4) + 3(a+2) - 2 = 3a^2 + 12a + 12 + 3a + 6 - 2 = 3a^2 + 15a + 16\).

Step 6: Compute \(\frac{f(a)}{2}\)

First find \(f(a) = 3a^2 + 3a - 2\). Then, dividing by 2, \(\frac{f(a)}{2} = \frac{3a^2 + 3a - 2}{2}\). The result is simply \(\frac{3a^2}{2} + \frac{3a}{2} - 1\).

Step 7: Evaluate \(f(2a)\)

Substitute \(x = 2a\) into the function \(f(x) = 3x^2 + 3x - 2\). Thus, \(f(2a) = 3(2a)^2 + 3(2a) - 2 = 3 \times 4a^2 + 6a - 2 = 12a^2 + 6a - 2\).

Step 8: Add \(f(a) + f(2)\)

Find \(f(a) = 3a^2 + 3a - 2\) and \(f(2) = 16\) from previous steps. Hence, \(f(a) + f(2) = 3a^2 + 3a - 2 + 16 = 3a^2 + 3a + 14\).

Step 9: Evaluate \(f(a+h)\)

Substitute \(x = a+h\) into the function \(f(x) = 3x^2 + 3x - 2\). Thus, \(f(a+h) = 3(a+h)^2 + 3(a+h) - 2\). Expanding \((a+h)^2 = a^2 + 2ah + h^2\), it becomes: \(3(a^2 + 2ah + h^2) + 3a + 3h - 2 = 3a^2 + 6ah + 3h^2 + 3a + 3h - 2\).

Key Concepts

Understanding Polynomials

A polynomial is a mathematical expression containing a sum of powers in one or more variables multiplied by coefficients. Polynomials are significant in mathematics because they represent a broad class of functions. In the exercise, we are given the polynomial function: \[f(x) = 3x^2 + 3x - 2\] This expression has three terms, each involving powers of the variable \(x\). Here’s a quick breakdown of this polynomial:

• The coefficient of \(x^2\) is 3, which means this polynomial is quadratic.
• The linear term is \(3x\). It affects the slope of the polynomial's graph.
• The constant term is -2, which shifts the entire graph up or down.

Understanding the structure of the polynomial is crucial for evaluating it using various methods, such as substitution.

Using the Substitution Method

The substitution method involves replacing a variable in an expression with a given value or another expression. This method is key to function evaluation. Let's see how substitution is applied in different cases from the exercise:

• For \(f(2)\), substitute \(x = 2\) in the function, resulting in a calculation of \(3(2)^2 + 3(2) - 2\).
• For \(f\left(\frac{2}{a}\right)\), substitute \(x = \frac{2}{a}\). This changes the entire expression, requiring careful adjustments with fractions.
• In \(f(a+h)\), use substitution to introduce a new variable \(h\), demonstrating how the function behaves as \(x\) shifts slightly from \(a\).

This method is powerful because it allows examination of the function's behavior at specific points or under modified conditions. It greatly aids in graphical interpretations and real-world applications.

Function Expressions Explained

Function expressions define relationships where each input has a corresponding output, typically expressed with variables like \(x\) and \(f(x)\). In this case, \[f(x) = 3x^2 + 3x - 2\] serves as a function expression indicating how each value of \(x\) is transformed through multiplication, addition, and subtraction. Here,

• \(f(x)\) reads as 'the function \(f\) of \(x\)', illustrating how \(x\) impacts the outcome.
• This form of notation is crucial for recognizing functional dependencies and applying operations systematically.
• The exercise requires evaluating these expressions at various points, like \(x = 2\) and \(x = -2\), to see how the output \(f(x)\) changes.

Understanding function expressions equips students with the foundation for more advanced topics like calculus, where functions may be composed or differentiated.

Simplification Techniques

Simplification is a process used to make expressions easier to interpret or compute by combining like terms or reducing complex fractions. In the exercise:

• When calculating \(2f(a)\), the original expression \(3a^2 + 3a - 2\) is multiplied by 2, leading to simplification in the expression: \(6a^2 + 6a - 4\).
• For \(\frac{f(a)}{2}\), the expression is divided by 2, simplifying each component: \(\frac{3a^2}{2} + \frac{3a}{2} - 1\).
• Expressions like \(f(a+h)\) are expanded and simplified through distribution of coefficients and combining like terms.

These techniques help manage algebraic expressions and make them manageable for problem-solving. Mastery of simplification builds a strong foundation for tackling complex mathematics efficiently.