Problem 9 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 9-16, complete the table by computing \(f(x)\) at the given values of \(x\). Use these results to estimate the indicated limit (if it exists). $$ \begin{array}{l} f(x)=x^{2}+1 ; \lim _{x \rightarrow 2} f(x) \\ \hline x \quad 1.9 \quad 1.99 \quad 1.999 \quad 2.001 \quad 2.01 \quad 2.1 \\ \hline \boldsymbol{f}(\boldsymbol{x}) & & & & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

Based on the given function \(f(x) = x^2 + 1\), we calculated the values of \(f(x)\) for the given input values, which allowed us to estimate the limit \(\lim_{x \rightarrow 2} f(x)\) to be \(5\).

Step 1: Calculate f(x) for each x value

We will compute the value of f(x) for each given x value: 1. \(f(1.9) = (1.9)^2 + 1\) 2. \(f(1.99) = (1.99)^2 + 1\) 3. \(f(1.999) = (1.999)^2 + 1\) 4. \(f(2.001) = (2.001)^2 + 1\) 5. \(f(2.01) = (2.01)^2 + 1\) 6. \(f(2.1) = (2.1)^2 + 1\)

Step 2: Compute the f(x) values

Now, we will compute the values for each f(x): 1. \(f(1.9) = 4.61\) 2. \(f(1.99) = 4.9601\) 3. \(f(1.999) = 4.996001\) 4. \(f(2.001) = 5.004001\) 5. \(f(2.01) = 5.0401\) 6. \(f(2.1) = 5.41\)

Step 3: Update the table with computed values

We will update the table with our computed values to get: \( \begin{array}{l} f(x)=x^{2}+1 ; \lim _{x \rightarrow 2} f(x) \\ \hline x \quad 1.9 \quad 1.99 \quad 1.999 \quad 2.001 \quad 2.01 \quad 2.1 \\ \hline \boldsymbol{f}(\boldsymbol{x}) \quad 4.61 \quad 4.9601 \quad 4.996001 \quad 5.004001 \quad 5.0401 \quad 5.41 \\ \hline \end{array} \)

Step 4: Estimation of the limit

By observing the values in the table, we see that as x approaches 2, f(x) gets closer to 5. Thus, we can estimate the limit \(\lim_{x \rightarrow 2} f(x)\) to be \(5\).

Key Concepts

Function Evaluation

When we talk about function evaluation, we're referring to figuring out the value of a function at specific points. In simple terms, it's like putting a number into a formula and seeing what you get out. For this exercise, the function we have is a simple polynomial: \[ f(x) = x^2 + 1 \] Our task is to evaluate this function at several predetermined points: 1.9, 1.99, 1.999, 2.001, 2.01, and 2.1. By substituting these values into our equation, we discover the output or the function value at each point:

• At \(x = 1.9\), \(f(1.9) = 1.9^2 + 1 = 4.61\)
• At \(x = 1.99\), \(f(1.99) = 1.99^2 + 1 = 4.9601\)
• At \(x = 1.999\), \(f(1.999) = 1.999^2 + 1 = 4.996001\)
• At \(x = 2.001\), \(f(2.001) = 2.001^2 + 1 = 5.004001\)
• At \(x = 2.01\), \(f(2.01) = 2.01^2 + 1 = 5.0401\)
• At \(x = 2.1\), \(f(2.1) = 2.1^2 + 1 = 5.41\)

This sort of mathematical process is crucial in helping us understand how functions behave at particular points.

Limit Estimation

Limit estimation involves finding the value that a function approaches as its input gets very close to a certain number. In the given exercise, we’re estimating the limit as \(x\) approaches 2 for the function \(f(x) = x^2 + 1\). To do this, we look at the values calculated in our function evaluation:

• For values approaching 2 from the left (1.9, 1.99, 1.999), \(f(x)\) gives us 4.61, 4.9601, and 4.996001. These values increase, approaching 5.
• For values approaching 2 from the right (2.001, 2.01, 2.1), \(f(x)\) gives us 5.004001, 5.0401, and 5.41. These values decrease, also getting closer to 5.

By observing both sequences getting closer to 5, we conclude that the estimated limit \(\lim_{x \to 2} f(x) = 5\). This method is extremely useful as it helps us understand the end behavior of functions around particular points, even if the function might not be defined exactly at those points.

Polynomial Functions

Polynomial functions form an essential part of calculus and algebra. They are expressions consisting of variables and coefficients, composed using only the operations of addition, subtraction, multiplication, and non-negative integer exponents. The given exercise involved a polynomial function: \[ f(x) = x^2 + 1 \] This specific function is a quadratic polynomial, as its highest power is 2. Polynomials like this are

• Continuous: You can draw them without lifting your pencil.
• Smooth: They have no sharp corners or breaks.
• Differentiable: You can find a slope (derivative) at every point.

Understanding polynomials is vital to grasp higher concepts in calculus, such as limits and derivatives because they exhibit predictable behavior. In this problem, we observe how the polynomial behaves around \(x = 2\), underlining that functions like these often allow us to easily estimate the limit.