Problem 51 - Single Variable Calculus

Question

Show that the function is continuous but not differentiable at the given value of \(x\). \(f(x)=|2 x-1| ; \quad x=\frac{1}{2}\)

Step-by-Step Solution

Answer

The function $f(x)=|2x-1|$ is continuous at $x=\frac{1}{2}$ since the left-hand limit and the right-hand limit both equal 0 (\(\lim_{x \to \frac{1}{2}^-} |2x-1|=0\) and \(\lim_{x \to \frac{1}{2}^+} |2x-1|=0\)). However, the function is not differentiable at this point because the derivative does not exist: $f'(x)=2$ for $x>\frac{1}{2}$ and $f'(x)=-2$ for $x<\frac{1}{2}$. Thus, the function is continuous but not differentiable at $x=\frac{1}{2}$.

Step 1: Evaluate the limit as x approaches the given value of x to test for continuity

We first need to find the left-hand limit and the right-hand limit of the function at \(x=\frac{1}{2}\). If both of these limits exist and are equal, then we can say that the function is continuous at \(x=\frac{1}{2}\). Left-hand limit: \[ \lim_{x \to \frac{1}{2}^-} |2x-1| \] Right-hand limit: \[ \lim_{x \to \frac{1}{2}^+} |2x-1| \]

Step 2: Determine the left-hand limit

Since we have the absolute value, we need to consider two cases. The function can be rewritten as: \[ f(x) = \begin{cases} 2x - 1, & x \geq \frac{1}{2} \\ 1 - 2x, & x < \frac{1}{2} \end{cases} \] Now, finding the left-hand limit: \[ \lim_{x \to \frac{1}{2}^-} |2x-1| = \lim_{x \to \frac{1}{2}^-}(1-2x) = 1-2\left(\frac{1}{2}\right) = 0 \]

Step 3: Determine the right-hand limit

Similarly, finding the right-hand limit: \[ \lim_{x \to \frac{1}{2}^+} |2x-1| = \lim_{x \to \frac{1}{2}^+}(2x-1)=2\left(\frac{1}{2}\right)-1 = 0 \] Since the left-hand limit and the right-hand limit both equal 0, we can conclude that the function is continuous at \(x=\frac{1}{2}\).

Step 4: Find the derivative of the function to check for differentiability

Now let's find the derivative of the function: \[ f'(x) = \begin{cases} 2, & x>\frac{1}{2} \\ -2, & x<\frac{1}{2} \end{cases} \] Notice that no derivative exists at the point \(x=\frac{1}{2}\), as the left and right sides of the function have different slopes.

Step 5: Determine differentiability

Since the derivative of the function does not exist at \(x=\frac{1}{2}\), we can conclude that the function is not differentiable at that point. Thus, we have shown that the function is continuous but not differentiable at \(x=\frac{1}{2}\).

Key Concepts

Limit of a Function

Understanding the limit of a function is crucial when analyzing its behavior at a specific point. The limit answers the question: What value does the function approach as the input, or x, gets closer to a certain number? This concept is foundational in calculus and allows us to deal with situations where a function does not return a clear output at a certain point.
A limit does not always exist, but if it does, it represents the value that the function outputs become arbitrarily close to as the input approaches a particular value. In the given exercise, we evaluate limits to ascertain the function's continuity at \(x = \frac{1}{2}\). If the function approaches the same value from both the left and the right at this point, the function is said to be continuous there, setting the stage for further analysis, like checking for differentiability.

Absolute Value Functions

Characteristics of Absolute Value Functions

Absolute value functions often introduce sharp turns or corners in their graphs at the point where the inside of the absolute value becomes zero. The function \(f(x) = |2x - 1|\) used in our exercise has such a characteristic at \(x = \frac{1}{2}\), which is why there is a possibility that it might not be differentiable at that point even though it's continuous.
In general, we split the analysis of an absolute value function into two parts or 'cases' - one for when the inside of the absolute value is positive and one for when it’s negative. This distinction is pivotal when assessing limits and derivatives.

Left-Hand and Right-Hand Limits

The concept of left-hand and right-hand limits dives into the behavior of function as it approaches a particular point from the left (\(x \to c^-\)) and right (\(x \to c^+\)) respectively. These unilateral limits are essential when a function has a discontinuity or a point at which its behavior suddenly changes — like a jump or a cusp.

• Left-Hand Limit: This evaluates what value the function is approaching as the input comes from values less than the point of interest.
• Right-Hand Limit: This evaluates what value the function is approaching as the input comes from values greater than the point of interest.

In our exercise, determining the left-hand and right-hand limits and finding them equal allowed us to confirm the function's continuity at \(x = \frac{1}{2}\).

Derivative of a Function

The derivative of a function at a point measures how the function value changes with a slight change in the input value; it's the function's rate of change or slope at that point. Not all functions have derivatives at all points. A function that has a sharp turn or cusp, such as with absolute value functions, may not be differentiable at that point because it lacks a singular, well-defined tangent.
In our exercise, the absolute value function \(f(x) = |2x - 1|\) has different slopes on the left and right of \(x = \frac{1}{2}\), leading to a 'jump' in the derivative. The left side approaches a slope of -2, while the right side approaches a slope of 2. Since these slopes don't match, there can be no single derivative at \(x = \frac{1}{2}\), thus the function is not differentiable there. This discrepancy in slopes essentially pinpoints the location of a 'non-smooth' point in the function's profile.