Problem 26

Question

Let $$ f(x)=\left\\{\begin{array}{cl} \frac{1}{x} & \text { for } x \geq 1 \\ 2 x+c & \text { for } x<1 \end{array}\right. $$ (a) Graph $f(x)$ when $c=0$, and determine whether $f(x)$ is continuous for this choice of $c$. (b) How must you choose $c$ so that $f(x)$ is continuous for all $x \in(-\infty, \infty)$ ?

Step-by-Step Solution

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Answer

For (a), the function is discontinuous for $ c = 0 $. For (b), choose $ c = -1 $ for continuity.

1Step 1: Understand the function

The function $ f(x) $ is a piecewise function that behaves differently based on the value of $ x $. For $ x \geq 1 $, the function is $ f(x) = \frac{1}{x} $. For $ x < 1 $, the function is $ f(x) = 2x + c $. The challenge is to determine the value of $ c $ that makes this function continuous.

2Step 2: Plot the function for c=0

For $ c = 0 $, the second part of the function becomes $ f(x) = 2x $ for $ x < 1 $. Thus, $ f(x) = \frac{1}{x} $ for $ x \geq 1 $ and $ f(x) = 2x $ for $ x < 1 $. Plot $ \frac{1}{x} $ from $ x = 1 $ onwards and $ 2x $ for $ x < 1 $. Notice that at $ x = 1 $, $ \frac{1}{1} = 1 $ and $ 2 \times 1 = 2 $, so the values do not match.

3Step 3: Determine continuity for c=0

The function $ f(x) $ is continuous if the left-hand limit ($ x \to 1^- $), right-hand limit ($ x \to 1^+ $), and $ f(1) $ are all equal. For $ x < 1 $, as $ x \to 1^- $, $ f(x) = 2x \to 2 \times 1 = 2 $. For $ x \to 1^+ $, $ f(x) = \frac{1}{x} \to \frac{1}{1} = 1 $. As these do not equal the function value itself at $ x=1 $, it is discontinuous for $ c=0 $.

4Step 4: Determine c for continuity

Set the limits from the left and right to equal $ f(1) $. Hence, $ 2 \cdot 1 + c = \frac{1}{1} $. Solving gives $ 2 + c = 1 $, thus $ c = 1 - 2 = -1 $.

5Step 5: Verify and conclude

With $ c = -1 $, the parts of the function are $ f(x) = \frac{1}{x} $ for $ x \geq 1 $ and $ f(x) = 2x - 1 $ for $ x < 1 $. At $ x = 1 $, both sides provide $ f(1) = 1 $. Therefore, the function becomes continuous for all $ x $.

Key Concepts

Continuity in CalculusGraphing FunctionsLimits and Continuity

Continuity in Calculus

Continuity in calculus is a fundamental concept that ensures a function behaves nicely, without any sudden jumps or gaps. We consider a function continuous at a point if three conditions are satisfied:

The function is defined at that point.
The limit of the function as it approaches the point from both sides exists.
The limit equals the function's value at that point.

For the function given in the exercise, which is piecewise, continuity means that at the boundary point, where the function definition changes (here at $x=1$), all these conditions must hold. We've found that by properly choosing $c$, in this case, $c=-1$, the function becomes continuous over its entire domain. This involves setting the left-hand limit $(x \to 1^-)$ and the right-hand limit $(x \to 1^+)$ equal to each other and to $f(1)$. Therefore, the close attention to limits and condition checks are instrumental in ensuring a function's continuity across changing limits and definitions.

Graphing Functions

Graphing functions helps to visualize the behavior of mathematical relationships. For piecewise functions, such as the one in the exercise, graphing involves understanding how each piece behaves, then connecting them correctly.
To graph the given function when $c=0$:

For $x \geq 1$, plot the curve $f(x) = \frac{1}{x}$. This is a hyperbolic curve that approaches the $x$-axis as $x$ increases.
For $x < 1$, plot the line $f(x) = 2x$. This is a linear function with a slope of 2, passing through the origin.

At $x=1$, these plots must connect seamlessly for the function to be continuous. For $c=0$, this fails because the two parts of the function meet at different values (1 and 2, respectively). By adjusting $c$ to $-1$, the linear part shifts down, aligning perfectly with the curve, creating a continuous graph. Graphical representation clarifies why continuity is not achieved just by original configurations, emphasizing the need for adjustments like modifying $c$ in this exercise.

Limits and Continuity

Limits play a pivotal role in determining a function's continuity. The limits of a function as it approaches a specific point from the left $(x \to 1^-)$ and from the right $(x \to 1^+)$ dictate if the function changes or jumps. For a function to be continuous at $x=1$, like in our piecewise example, the following must occur:

Both the left-hand and right-hand limits must exist.
The limits from either direction must be equal.
This common limit must match the actual function value at $x=1$.

In our exercise, the initial choice of $c=0$ resulted in different left and right limits at $x=1$, meaning the function was not continuous. By solving the equation $2+c = 1$, and adjusting $c$ to $-1$, both sides aligned at $f(1) = 1$, ensuring that both the limits and the function value were consistent. This adjustment ensures the entire function is seamless and continuous, showcasing the interconnectedness of limits and function continuity in calculus.

Problem 26

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