Problem 4 - Calculus for Biology and Medicine

Question

In Problems 1-4, show that each function is continuous at the given value. $$ f(x)=x^{2}+1, c=-1 $$

Step-by-Step Solution

Answer

Function \( f(x) = x^2 + 1 \) is continuous at \( x = -1 \).

Step 1: Understand the Definition of Continuity

A function \( f(x) \) is continuous at a point \( c \) if three conditions are met: (1) \( f(c) \) is defined, (2) \( \lim_{{x \to c}} f(x) \) exists, and (3) \( \lim_{{x \to c}} f(x) = f(c) \). We need to verify these conditions for the function \( f(x) = x^2 + 1 \) at \( c = -1 \).

Step 2: Check Whether the Function is Defined at \( c \)

Calculate \( f(-1) \). Substitute \( x = -1 \) into the function: \[ f(-1) = (-1)^2 + 1 = 1 + 1 = 2 \]. Since \( f(-1) = 2 \), \( f(x) \) is defined at \( x = -1 \).

Step 3: Calculate the Limit of the Function as \( x \) Approaches \( c \)

Now, find the limit \( \lim_{{x \to -1}} f(x) \) using the same expression: \[ \lim_{{x \to -1}} (x^2 + 1) = (-1)^2 + 1 = 2 \]. This shows that the limit exists and is equal to 2.

Step 4: Compare the Limit to the Function Value at \( c \)

Now confirm that \( \lim_{{x \to -1}} f(x) = f(-1) \). We found that both the limit and the function value are 2, i.e., \[ \lim_{{x \to -1}} f(x) = 2 = f(-1) \].

Step 5: Conclusion on Continuity

Since all conditions of continuity are satisfied: \( f(-1) \) is defined, \( \lim_{{x \to -1}} f(x) \) exists and equals 2, and \( \lim_{{x \to -1}} f(x) = f(-1) \), the function \( f(x) = x^2 + 1 \) is continuous at \( x = -1 \).

Key Concepts

Limit of a Function

The concept of the limit of a function is essential in calculus and helps us understand the behavior of functions as the input approaches a specific value. In simpler terms, when we talk about the limit of a function, we are asking what happens to the value of the function as we get closer and closer to a certain point on the x-axis.
To determine the limit of a function, say for the function \( f(x) = x^2 + 1 \) as \( x \) approaches \( -1 \), we calculate how \( f(x) \) behaves as \( x \) gets very close to \( -1 \). We substitute values closer and closer to \( -1 \) and observe the trend of the function's output. If our calculated limit approaches a real number, we say that the limit exists.
For the given exercise, the limit was found to be 2 when \( x \) approaches \( -1 \). This tells us that as \( x \) gets infinitesimally close to \( -1 \), the value of \( f(x) \) nears 2. It is crucial for the limit to exist to consider whether a function is continuous at a point.

Function Evaluation

Function evaluation is the process through which we find the output of a function for a specific input value. This is typically done by substituting the input value into the function's formula.
For the function \( f(x) = x^2 + 1 \), evaluating the function at \( x = -1 \) means substituting \( -1 \) into the function, resulting in:

• \( f(-1) = (-1)^2 + 1 \)
• \( f(-1) = 1 + 1 \)
• \( f(-1) = 2 \)

This calculation shows us that the function's output at \( x = -1 \) is 2. Moreover, verifying that the function is defined at \( x = -1 \) is crucial in determining continuity. Function evaluation is a basic yet important task that allows us to explore many mathematical properties, including checking for continuity, as seen in this exercise.

Polynomial Functions

Polynomial functions are sums of terms consisting of a variable raised to an exponent and multiplied by a coefficient. These functions are of the form \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \, ... \, + a_1 x + a_0 \). For our function, \( f(x) = x^2 + 1 \), it is a simple polynomial of degree 2, often referred to as a quadratic function.
Polynomial functions, like \( f(x) = x^2 + 1 \), have many interesting properties:

• They are smooth and continuous over all real numbers.
• They do not have any gaps, jumps, or infinite oscillations.
• They have a predictable behavior and can easily be evaluated for derivatives and integrals.

The function \( f(x) = x^2 + 1 \) is continuous at any value of \( x \), including \( x = -1 \), as verified in the original exercise. This takes us to the broader concept of continuity for polynomial functions, which is guaranteed by their inherent construction and properties.