Problem 13 - Calculus

Question

In Problems \(1-20\), find the numbers \(c\) that make \(f(x)\) into (A) a continuous function and (B) a differentiable function. In one case \(f(x) \rightarrow f(a)\) at every point, in the other case \(\Delta f / \Delta x\) has a limit at every point. $$ f(x)=\left\\{\begin{array}{cl} \frac{x^{2}+c}{x-1} & x \neq 1 \\ 2 & x=1 \end{array}\right. $$

Step-by-Step Solution

Answer

The value of \( c \) that makes \( f(x) \) continuous and differentiable at \( x = 1 \) is \( -1 \).

Step 1: Understand the Problem Statement

To solve the problem, we need to find the value of \( c \) such that \( f(x) \) is both continuous and differentiable at \( x = 1 \). This means first ensuring the function is continuous at \( x = 1 \), and second ensuring it is differentiable at the same point. We will use limits to find appropriate conditions for continuity and differentiability.

Step 2: Check Continuity

A function \( f(x) \) is continuous at \( x = a \) if \( \lim_{x \to a} f(x) = f(a) \). In this case, we have \( f(1) = 2 \). Thus, to ensure continuity at \( x = 1 \), we need \( \lim_{x \to 1} \frac{x^2 + c}{x - 1} = 2 \).

Step 3: Calculate the Limit for Continuity

To find \( \lim_{x \to 1} \frac{x^2 + c}{x - 1} \), substitute \( x = 1 + h \) and let \( h \to 0 \): \[ \lim_{h \to 0} \frac{(1+h)^2 + c}{h} = \lim_{h \to 0} \frac{1 + 2h + h^2 + c}{h}. \] This results in: \[ \lim_{h \to 0} \left( \frac{1 + c}{h} + 2 + h \right). \] For this limit to be 2, \( \frac{1 + c}{h} \to 0 \) as \( h \to 0 \), implying \( 1 + c = 0 \) or \( c = -1 \).

Step 4: Check Differentiability

A function \( f(x) \) is differentiable at \( x = a \) if \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) exists. We computed \( f(1) = 2 \) and the limit for the change in function is already established as a constant: \( \lim_{h \to 0} \left( \frac{1 + c}{h} + 2 + h \right) = 2 \). Since \( \frac{1 + c}{h} \to 0 \) implies \( c = -1 \), the expression becomes zero, ensuring differentiability.

Step 5: Final Conclusion

For the function \( f(x) \) to be both continuous and differentiable at \( x=1 \), \( c \) must be \(-1\). This ensures the limit equals the function value and that the slope between points tends to zero, thus exhibiting differentiability.

Key Concepts

Limits

Limits are fundamental in calculus and are essential for defining both continuous and differentiable functions. When discussing limits, we refer to the value that a function approaches as the input gets arbitrarily close to a specific point. In our exercise, we focused on ensuring that

• the limit of the function as it approaches a certain point is equal to the function's value at that point
• which is key for continuity.

This involves evaluating expressions like \( \lim_{x \to 1} \frac{x^2 + c}{x - 1} \).
By substituting variables and calculating the limit, we find that in order for the function to be continuous at \( x = 1 \), it must match the given function value. Understanding limits allows us to determine necessary conditions for a function to be smooth at specific points, which is particularly relevant when calculating derivatives impacted by the same principles.

Continuous Functions

A function is considered continuous at a point \( x = a \) if the limit of the function as \( x \) approaches \( a \) equals the function's value at \( a \). In simpler terms, there are no sudden jumps or gaps in the function at that point. For this condition, we use the formula
\( \lim_{x \to a} f(x) = f(a) \).
In our example, since \( f(1) = 2 \), the task was to ensure that \( \lim_{x \to 1} \frac{x^2 + c}{x - 1} = 2 \).
This means plugging \( x = 1 \) into \( f(x) \) seamlessly results in the intended output of \( 2 \) without any undefined behavior or breaks.
Calculating these stages proves crucial in ensuring all values are interconnected without disruptions, making a function continuous at that point.

Differentiable Functions

Differentiability refers to a function having a well-defined tangent line at a specific point, which means that it doesn't have any sharp corners or cusps. A function is differentiable at \( x = a \) if there exists:
\( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)
and the limit has a numerical value.
This effectively means that as you zoom in infinitely close to \( x = a \), the function appears as a straight line. In terms of our exercise, once we assured the function is continuous at \( x = 1 \) by selecting \( c = -1 \), we achieved differentiability because the potential discontinuities were resolved.
The function smoothly transitions through the point, ensuring the slope of the tangent line exists and is consistent, as the expression resolves to a constant. Understanding this concept ensures students grasp the rounded nature of differentiable functions, as opposed to erratically changing functions.