Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the function \(f\) is differentiable for all values of \(x,\) then \(f\) is continuous for all values of \(x\) b. The function \(f(x)=|x+1|\) is continuous for all \(x\), but not differentiable for all \(x\). c. It is possible for the domain of \(f\) to be \((a, b)\) and the domain of \(f^{\prime}\) to be \([a, b]\)

Short answer

Provide an explanation or a counterexample. Answer: Yes, it is possible for a function to be continuous for all x but not differentiable for all x. A counterexample would be the function \(f(x) = |x+1|\), which is continuous for all values of \(x\) but not differentiable when \(x=-1\). At \(x=-1\), there is a sudden change in the slope of the function, causing the left and right derivatives to be unequal, and hence, the derivative does not exist at that point.

Step-by-step solution

Statement a: Differentiability implies continuity

If a function is differentiable at a point, it means that the derivative exists at that point, and therefore, the function is also continuous at that point. More formally, if \(f\) is differentiable for all values of \(x\), then \(f'(x)\) exists for all values of \(x\). Since the existence of the derivative implies continuity, the statement holds true. So, if the function \(f\) is differentiable for all values of \(x\), then \(f\) is continuous for all values of \(x\).

Statement b: Analyzing the function \(f(x)=|x+1|\)

The given function is \(f(x)=|x+1|\). Recall that an absolute function has a "kink" or an abrupt change in direction at the point where the argument equals zero. In this case, the function is continuous for all values of \(x\), as there are no breaks or jumps. However, the function is not differentiable when \(x=-1\) because there is a sudden change in the slope of the function at that point. To be more specific, the left derivative and right derivative are not equal at this point, meaning the derivative does not exist at \(x=-1\). Therefore, this statement is true - the function is continuous for all \(x\), but not differentiable for all \(x\).

Statement c: Comparing domains of \(f\) and \(f'\)

The statement says the domain of \(f\) is \((a, b)\), which means the function is defined for all values of \(x\) strictly between \(a\) and \(b\). The statement also says the domain of \(f'\) is \([a, b]\), which means the derivative of \(f\) exists for all values of \(x\) in the closed interval \([a, b]\). However, it is not possible for a function's derivative to exist at an endpoint where the function itself is not defined. So, this statement is false. It is not possible for the domain of \(f\) to be \((a, b)\) while the domain of \(f'\) is \([a, b]\).

Key concepts

Differentiable Functions

Understanding the idea of differentiable functions is fundamental in calculus. Put simply, for a function to be differentiable at a point, it must have a well-defined tangent at that point, which means it’s smooth without any breaks, bends, or sharp corners. The concept of differentiability is crucial because it ties directly into the derivative of a function — a measure of how the output of the function changes as its input changes.

Let's look at an example to clarify this concept. If we consider the function \( f(x) = x^2 \), it is differentiable everywhere because its graph is a smooth parabola, with a clear slope at every point. When we differentiate \( f(x) = x^2 \), we get \( f'(x) = 2x \), which exists for all values of \( x \).

However, not all functions are smooth and continuous. For instance, a function with a sharp corner or cusp is not differentiable at that point because you can’t neatly draw a single tangent line there. A clear understanding of this concept helps us to predict the behavior of functions and to grasp the underlying mechanics of calculus.

Continuous Functions

A continuous function, in its simplest terms, is one where small changes in the input lead to small changes in the output. There aren’t any abrupt jumps or breaks in the graph of the function. For example, the function \( g(x) = x^3 \) is continuous since it smoothly flows without interruption for all real numbers. Intuitively, you can draw the graph of a continuous function without lifting your pencil from the paper.

A function being continuous at a point means the function’s value at that point equals the limit of the function as it approaches the point. Mathematically, a function \(f\) is continuous at a point \(c\) if \( \text{lim}_{x \to c} f(x) = f(c) \). What’s especially interesting about continuous functions is that if a function is continuous over an interval, you’re assured that it fills in the entire space between the start and endpoint — there will be no gaps.

Absolute Value Function Continuity

The absolute value function, denoted by \( f(x) = |x| \), is continuous everywhere, but it presents a unique twist when it comes to differentiability. This is due to its V-shaped graph, which has a sharp point at the origin (where \( x = 0 \)). While this function does not have any breaks or gaps (hence it’s continuous), it has a problem when you try to determine its slope at the vertex of the ‘V’. The slope abruptly changes from positive to negative there, which means there’s no single tangent line that can be drawn at that point.

Moreover, the provided exercise further explores this concept using the function \( f(x) = |x+1| \), illustrating that even though the function is continuous for all real numbers \(x\), it fails the differentiability test at \( x = -1 \) because of the change in direction. Visualizing the graph helps understand this concept: it is smooth and connected, yet the corner at \( x = -1 \) means the slope isn't consistent from both sides.

Domain of a Function

The domain of a function is the complete set of input values (often \(x\)-values) for which the function is defined. Think of it as the ‘home’ of all possible inputs where the function can comfortably operate without running into any issues like division by zero or taking the square root of a negative number.

For example, the domain of \( h(x) = \frac{1}{x} \) does not include zero, because division by zero is undefined. Thus, the domain is all real numbers except zero. In the context of the original exercise, it’s important to note that while a function can be defined on an open interval \((a, b)\), meaning it exists between but not including \(a\) and \(b\), the domain of its derivative might be different because the derivative might not exist at the boundaries of the interval.

This nuanced understanding of domain helps students in graphing functions and finding the range of possible outputs. Knowing precisely where a function lives, in terms of input values, is foundational in the study of calculus and mathematical analysis.