Problem 24 - Algebra 2 and Trigonometry

Question

In \(19-28 :\) a. Find \(\mathrm{f}(a)\) for each given function. b. Is \(a\) a root of the function? $$ f(x)=2 x^{3}+3 x^{2}-1 \text { and } a=\frac{1}{2} $$

Step-by-Step Solution

Answer

\( f(\frac{1}{2}) = 0 \) and \( \frac{1}{2} \) is a root of the function.

Step 1: Substitute 'a' into the function

To find \( f(a) \), we need to substitute \( a = \frac{1}{2} \) into the function. This means replacing all \( x \) in \( f(x) = 2x^3 + 3x^2 - 1 \) with \( \frac{1}{2} \).

Step 2: Calculate \( f(\frac{1}{2}) \)

Start by calculating each part of the equation separately:- First, calculate \((\frac{1}{2})^3 = \frac{1}{8}\).- Then calculate \(2 \times \frac{1}{8} = \frac{1}{4}\).- Next, calculate \((\frac{1}{2})^2 = \frac{1}{4}\).- Then calculate \(3 \times \frac{1}{4} = \frac{3}{4}\).- Now substitute these results back into the equation: \(f(\frac{1}{2}) = \frac{1}{4} + \frac{3}{4} - 1\).

Step 3: Simplify the expression

Combine the fractions and constants in the expression:- \( \frac{1}{4} + \frac{3}{4} = 1\)- Then, subtract 1 from this sum: \( 1 - 1 = 0\).

Step 4: Determine if 'a' is a root

A value \( a \) is a root of the function if \( f(a) = 0 \).- In this case, \( f(\frac{1}{2}) = 0 \), so \( a = \frac{1}{2} \) is indeed a root of the function.

Key Concepts

Roots of Polynomial

Polynomials are an essential concept in algebra and calculus with various applications. When we talk about the roots of a polynomial, we refer to the values of the variable that make the entire polynomial equal to zero. Finding these roots can be seen as asking where a curve touches or crosses the x-axis on a graph.

To determine if a given value is a root, we substitute it into the polynomial equation and see if the result is zero. If it is, then the value is indeed a root. In our example, we checked if \( a = \frac{1}{2} \) was a root of the polynomial \( f(x) = 2x^3 + 3x^2 - 1 \) by plugging \( a \) into the equation:

• After evaluating the polynomial \( f(\frac{1}{2}) \), we found that the output was zero: \( f(\frac{1}{2}) = 0 \).

This confirms that \( \frac{1}{2} \) is a root. Finding roots can often involve factoring, using the quadratic formula for quadratics, or applying numerical methods like the Newton-Raphson method for more complicated polynomials.

Function Evaluation

Evaluating functions is a fundamental concept in mathematics. It involves finding the output of a function for a given input. When we talk about function evaluation, we're substituting a specific value for the variable and computing the results according to the function's formula.

In the context of this problem, we evaluated the function \( f(x) = 2x^3 + 3x^2 - 1 \) at \( a = \frac{1}{2} \). This means that we replaced every \( x \) in the function with \( \frac{1}{2} \) and calculated the result.

• The operation requires careful attention to each term: computing powers, performing multiplication, then simplifying & combining terms.
• The result of \( f(\frac{1}{2}) \) turned out to be 0, as seen from the evaluation steps.

Evaluating functions accurately is key in many areas of math, from basic algebra to advanced calculus, as it allows for the analysis and understanding of the behavior of functions under various conditions.

Substitution in Functions

Substitution is a basic yet powerful tool in mathematics that allows us to simplify complex expressions and solve equations by replacing variables with specific numerical values. It is particularly useful in evaluating polynomial functions, where variables appear multiple times, in multiple terms.

In our worked example, we performed substitution by replacing every instance of \( x \) in the function \( f(x) = 2x^3 + 3x^2 - 1 \) with \( \frac{1}{2} \). Here's a quick overview of what we did:

• Calculated \( (\frac{1}{2})^3 = \frac{1}{8} \) and multiplied it by 2 to get \( \frac{1}{4} \).
• Determined \( (\frac{1}{2})^2 = \frac{1}{4} \) and multiplied this by 3 to achieve \( \frac{3}{4} \).
• Added the results and subtracted 1, getting zero in the end.

Substitution allows for a clear pathway to evaluate the function and decide whether a particular value is a root, as each substituted value provides a snapshot of the function's behavior at that point.