Problem 41 - University Calculus: Early Transcendentals

Question

Determine if the piecewise-defined function is differentiable at the origin. $$f(x)=\left\\{\begin{array}{ll}2 x-1, & x \geq 0 \\ x^{2}+2 x+7, & x<0\end{array}\right.$$

Step-by-Step Solution

Answer

The function is not differentiable at the origin because it is not continuous at that point.

Step 1: Verify Continuity at the Origin

To check if the function is differentiable at the origin, it must first be continuous there. Calculate the limit of the function from the left and right at the origin and ensure both limits are equal to the function's value at zero. \[ f(0) = 2(0) - 1 = -1 \] Check the left limit: \[ \lim_{x \to 0^-} (x^2 + 2x + 7) = 0^2 + 2(0) + 7 = 7 \] Check the right limit: \[ \lim_{x \to 0^+} (2x - 1) = 2(0) - 1 = -1 \] Since the left limit and the right limit are not equal, the function is not continuous at zero.

Step 2: Evaluate Differentiability

For differentiability at a point, a function must first meet the requirement of continuity. Since Step 1 established that the function is not continuous at zero, it cannot be differentiable at zero either.

Key Concepts

Continuity

Understanding continuity is essential when analyzing functions, especially when determining differentiability. A function is said to be continuous at a point if there is no interruption or abrupt change at that point.
To check continuity at a specific point, like zero in this case, we need to verify three conditions:

• The function needs to be defined at that point.
• The limit of the function as it approaches the point from the left (left-hand limit) should equal the limit as it approaches from the right (right-hand limit).
• These limits should be equivalent to the function's value at the point.

This means, for the piecewise function provided, verifying continuity involves comparing the limits from both directions. In our exercise, as calculated:

• The left-hand limit at zero was 7.
• The value of the function at zero was -1.
• The right-hand limit was also -1.

Since the left limit and the function's value at the origin are not the same, the function is not continuous at zero. Simple as that!

Piecewise Functions

Piecewise functions are a fascinating aspect of calculus. Unlike single-expression functions, piecewise functions consist of different expressions for different intervals of the input variable.
They are appealing because they can model more complex real-world scenarios, like our function above.

• For instance, in our exercise, one expression applies for non-negative values of x, while another applies for negative ones.
• These functions are typically defined using different expressions based on certain conditions.

When analyzing these functions, crucial insights into their behavior come from examining transitions between their pieces. Here, understanding these transitions at a specific point, such as zero, helps determine if there is continuity or differentiability there.
The behavior of the separate components can drastically affect the overall function's properties, as noticeable in the exercise where the two component expressions did not align seamlessly at zero.

Limits

Limits allow us to explore what happens to a function as it approaches a certain point. They're crucial in the continuity and differentiability discussions.
In our exercise, limits helped determine the behavior of the piecewise function at zero.
The process involves viewing how the output changes as inputs get closer to a particular value.

• Consider the left-hand limit: approaching zero from the negative side, we used the expression for x < 0.
• For the right-hand limit, we approached zero from the positive side using the expression for x ≥ 0.

Calculating limits for both sides revealed the disparity: the left-hand limit was 7, while the right-hand limit was -1, matching the function's value at zero.
Because these limits differ, it illustrates the essential point that continuity—and thereby differentiability—doesn't hold at zero.