Question

Prove that the following functions donot have an antiderivative on any interval which contains \(0 .\) (i) \(y=\left\\{\begin{array}{cl}x+1, & x>0 \\ x, & x \leq 0\end{array}\right.\) (ii) \(y= \begin{cases}x^{2}, & x \neq 0 \\ 1, & x=0\end{cases}\)

Short answer

Question: Prove that the following functions do not have antiderivatives on any interval containing 0: 1. \(y = x\) if \(x \leq 0\) and \(y = x + 1\) if \(x > 0\) 2. \(y = x^2\) if \(x \neq 0\) and \(y = 1\) if \(x = 0\) Solution: We examined each function and found that they are both discontinuous at \(x=0\). In the first function, the left and the right limits as \(x\) approaches 0 are different (0 and 1, respectively). In the second function, the limit of the function as \(x\) approaches 0 is 0, but the function is defined to be 1 at \(x=0\). Since the functions are not continuous, they do not have antiderivatives on any interval containing 0.

Step-by-step solution

(Function 1: Analyzing the intervals)

For the first function, we have two intervals to consider: \(x > 0\) and \(x \leq 0\). Let's examine the function, \(y\), and its limits as \(x\) approaches 0 from both the left and right sides.

(Function 1: Limit from the left)

As \(x\) approaches 0 from the left side, we have the function \(y = x, \, x \leq 0\). The limit as \(x\) approaches 0 from the left is: $$\lim_{x \to 0^-} y = \lim_{x \to 0^-} x = 0$$

(Function 1: Limit from the right)

As \(x\) approaches 0 from the right side, we have the function \(y = x + 1, \, x > 0\). The limit as \(x\) approaches 0 from the right is: $$\lim_{x \to 0^+} y = \lim_{x \to 0^+} (x+1) = 1$$ Since the left and right limits are not equal, \(y\) is not continuous at \(x=0\), and therefore does not have an antiderivative on any interval containing 0.

(Function 2: Analyzing the intervals)

For the second function, we have two intervals to consider: \(x \neq 0\) and \(x = 0\). Let's examine the function, \(y\), and its limits as \(x\) approaches 0 from both the left and right sides.

(Function 2: Limit from the left and right)

As \(x\) approaches 0 from both the left and right sides, we have the function \(y = x^2, \, x \neq 0\). The limit as \(x\) approaches 0 from both sides is: $$\lim_{x \to 0} y = \lim_{x \to 0} x^2 = 0$$ However, at x=0, the function is defined to be \(y=1\). Since the limit does not match the value of the function at \(x=0\), \(y\) is not continuous at \(x=0\), and therefore does not have an antiderivative on any interval containing 0.

Key concepts

Piecewise Functions Continuity

Understanding the continuity of piecewise functions is crucial in calculus, especially when dealing with antiderivatives. A piecewise function is a function that is defined by multiple sub-functions, each applying to a certain interval of the domain.

Continuity of piecewise functions, like any function, is necessary for the existence of an antiderivative. For a function to be continuous at a point, it must satisfy three conditions: the function must be defined at that point, the limit as the variable approaches the point must exist, and the limit must equal the value of the function at that point.

However, when there is a discontinuity, such as a jump discontinuity at the point where the sub-functions meet, which can often happen with piecewise functions, the function does not satisfy these conditions. As a result, an antiderivative cannot exist at that point, affecting the intervals around it.

Limit of a Function

The limit of a function as the variable approaches a particular value is a foundational concept in calculus. It describes the behavior of the function as the variable gets very close to a certain number, but does not necessarily include the value at that number.

In the context of antiderivatives, the existence of a limit at every point in an interval is essential. If the limits from the left and right do not match at a point, the function exhibits what is called a jump discontinuity. It results in a break in the function, which can be visualized as a 'jump' on the graph. For the existence of antiderivatives, the function must not have any such discontinuities within the interval of interest, which is why limits are so closely examined when discussing antiderivatives.

Existence of Antiderivatives

Antiderivatives, or indefinite integrals, are the reverse of derivatives. They are the functions that can give us back the original function when we differentiate them. Not every function has an antiderivative, which can be tied to its continuity properties.

An antiderivative exists only if the function is continuous over the interval where the antiderivative is being sought. This is due to the Fundamental Theorem of Calculus, which intimately links the process of taking antiderivatives (integration) to the area under the curve of a function. If a function has any type of discontinuity within an interval, such as leap, jump, or infinite discontinuity, it cannot have an antiderivative across that interval. This stems from the fact that the 'area under the curve' concept breaks down at points of discontinuity.

Discontinuity in Functions

Discontinuity in functions refers to points where a function is not continuous. There are several types of discontinuities, with common types being point, jump, and infinite discontinuities. Point discontinuity occurs when a function is undefined at a single point or the limit does not match the function's value. Jump discontinuity, as seen in piecewise functions, occurs when there is a sudden change in the function's value. Infinite discontinuity happens when the function goes to infinity at a point.

It's important for students to recognize discontinuities as they impact the differentiability and integrability of a function. In particular, if a function has a discontinuity at a point, it cannot have a derivative at that point, and its antiderivative may not exist across intervals including that point. Hence, identifying points of discontinuity is a key step in solving calculus problems related to differentiation and integration.