Problem 11 - Calculus with Concepts in Calculus

Question

Find the numerical value of the function at the given values of \(a\). $$ f(x)=\frac{-2}{169} x^{2}+\frac{4}{13} x+3 ; a=3.2,25.5 $$

Step-by-Step Solution

Answer

The function evaluates to 3.864 at \( a = 3.2 \) and 3.153 at \( a = 25.5 \).

Step 1: Substitute a = 3.2 into Function

The first given value for \( a \) is 3.2. To evaluate the function, substitute \( x = 3.2 \) into the function \( f(x) = \frac{-2}{169}x^2 + \frac{4}{13}x + 3 \). This gives: \[ f(3.2) = \frac{-2}{169}(3.2)^2 + \frac{4}{13}(3.2) + 3 \]

Step 2: Calculate f(3.2)

First, compute \((3.2)^2\), which is 10.24. Substitute this value into the function equation:\[ f(3.2) = \frac{-2}{169} \times 10.24 + \frac{4}{13} \times 3.2 + 3 \] Breaking this down:1. \(\frac{-2}{169} \times 10.24 \approx -0.121 \) (rounded to three decimal places)2. \(\frac{4}{13} \times 3.2 = \frac{12.8}{13} \approx 0.985 \) (rounded to three decimal places)3. Adding these with 3 gives:\[ -0.121 + 0.985 + 3 = 3.864 \]

Step 3: Substitute a = 25.5 into Function

The second given value for \( a \) is 25.5. To evaluate the function, substitute \( x = 25.5 \) into the function \( f(x) = \frac{-2}{169}x^2 + \frac{4}{13}x + 3 \). This gives: \[ f(25.5) = \frac{-2}{169}(25.5)^2 + \frac{4}{13}(25.5) + 3 \]

Step 4: Calculate f(25.5)

First, compute \((25.5)^2\), which is 650.25. Substitute this value into the function equation:\[ f(25.5) = \frac{-2}{169} \times 650.25 + \frac{4}{13} \times 25.5 + 3 \]Breaking this down:1. \(\frac{-2}{169} \times 650.25 \approx -7.693 \) (rounded to three decimal places)2. \(\frac{4}{13} \times 25.5 = \frac{102}{13} \approx 7.846 \) (rounded to three decimal places)3. Adding these with 3 gives:\[ -7.693 + 7.846 + 3 = 3.153 \]

Key Concepts

Polynomial Functions

Polynomial functions are expressions involving variables raised to whole number powers, combined using addition, subtraction, and multiplication. They can be as simple as a linear function like \(f(x) = 2x + 1\), or much more complex with higher degree terms and coefficients. For instance, the function \(f(x) = \frac{-2}{169}x^2 + \frac{4}{13}x + 3\) is a quadratic polynomial because its highest power of \(x\) is 2.

Every polynomial has a degree, which indicates the highest exponent of the variable in the expression. In our example, it's degree 2. The coefficients in a polynomial are the numbers before the variables, and they can be fractions, whole numbers, or decimals. The constant term is the term without a variable, which in \(f(x)\) is 3.

• Quadratic polynomial: Power of 2, as seen in \(x^2\)
• Coefficients: Typically rational numbers, such as \(-\frac{2}{169}\) and \(\frac{4}{13}\)
• Constant: The unchanging numeric part, like \(3\)

Understanding polynomial functions is essential for solving problems involving curves and paths in mathematics and real-world situations.

Substitution Method

The substitution method is a straightforward process used to evaluate functions by replacing the variable with a given number. This method helps to find the output of a function for specific inputs. It's a fundamental technique in solving and simplifying polynomial expressions.

When you are given a function, like \(f(x) = \frac{-2}{169}x^2 + \frac{4}{13}x + 3\), and an input value, such as \(a = 3.2\), you're asked to substitute \(x = 3.2\) into the function. This means replacing every \(x\) in the expression with 3.2. After substitution, the polynomial expression becomes a straightforward arithmetic problem.

• Replace each occurrence of the variable \(x\) with the number given
• Ensure operations follow the order: exponentiation, multiplication/division, and addition/subtraction
• Result in a single numerical value as the output

This method is reliable and effective, especially for polynomial functions, ensuring accurate function values by evaluating step by step.

Function Evaluation

Evaluating a function involves substituting a specific input value into the function and performing the arithmetic to find the result. In the context of polynomial functions, function evaluation helps us understand how these expressions behave with different inputs.

For instance, evaluating \(f(x) = \frac{-2}{169}x^2 + \frac{4}{13}x + 3\) at \(x=3.2\), requires substituting \(3.2\) into the equation. After substituting, calculate each part of the polynomial:
1. Compute \(x^2\), which for 3.2 is \(10.24\).2. Solve each term by multiplication with the coefficients.3. Add the constant term to the results of other calculations.

This approach will yield the function value for any given input. It's essential for graphing polynomial functions too, as it reveals how the graph will behave at each point.

• Calculate terms influ99enced by the space and order of operations
• Result in interpreting real-life scenarios or mathematical problems

By consistently applying this technique, mastering function evaluation becomes much simpler and allows better insight into the function's properties and graph shape.