Question

In Exercises \(11-18,\) use the function \(f\) defined and graphed below to answer the questions. $$f(x)=\left\\{\begin{array}{ll}{x^{2}-1,} & {-1 \leq x<0} \\\ {2 x,} & {0 < x < 1} \\ {1,} & {x=1} \\ {-2 x+4,} & {1 < x < 2} \\ {0,} & {2 < x < 3}\end{array}\right.$$ (a) Does \(f(1)\) exist? (b) Does \(\lim _{x \rightarrow 1} f(x)\) exist? (c) Does \(\lim _{x \rightarrow 1} f(x)=f(1) ?\) (d) Is \(f\) continuous at \(x=1 ?\)

Short answer

(a) Yes, \(f(1)\) exists and is equal to 1. (b) Yes, \(\lim _{x \rightarrow 1} f(x)\) exists and is equal to 2. (c) No, \(\lim _{x \rightarrow 1} f(x)\) does not equal to \(f(1)\). (d) No, \(f\) is not continuous at \(x = 1\).

Step-by-step solution

Determine existence of function value

We look at the value of \(f(1)\). According to the given function definition, \(f(1) = 1. \) Thus, \(f(1)\) exists and equals 1.

Determine existence of the limit

The limit at \(x = 1\) is defined if and only if the left-hand limit (\(\lim _{x \rightarrow 1^-} f(x)\)) equals to right-hand limit (\(\lim _{x \rightarrow 1^+} f(x)\)). From the function definition, the left-hand limit \( \lim _{x \rightarrow 1^-} f(x) = 2*1 = 2\) and right-hand limit \( \lim _{x \rightarrow 1^+} f(x) = -2*1 + 4= 2 \). As both limits are equal, hence, \(\lim _{x \rightarrow 1} f(x)\) exists and is equal to 2.

Compare the limit with the function value

Comparing our obtained values, we see that the function value at \(x = 1\) (which equals to 1) is not equal to the limit at \(x = 1\) (which is 2). Therefore, \(\lim _{x \rightarrow 1} f(x)\) does not equal to \(f(1). \)

Determine Function Continuity

A function is continuous at a point if and only if the function is defined at the point, the limit at the point exists, and the function value equals to the limit at the point. In this case, we find that the function is defined at \(x = 1\), and the limit at \(x = 1\) exists as well, but \(f(1) \) is not equal to the \(\lim _{x \rightarrow 1} f(x) \). Thus, \(f\) is not continuous at \(x = 1.\)

Key concepts

Piecewise Functions

Piecewise functions are mathematical expressions defined by different formulas for different intervals of the input variable, usually denoted by 'x'. These types of functions are like a patchwork quilt: each piece has its own pattern, but together they create a complete intricate design.

Consider the provided exercise where the function f(x) represents a piecewise function with different expressions depending on the interval to which 'x' belongs. For example, the expression x^2 - 1 defines f(x) when -1 ≤ x < 0, while the expression 2x defines f(x) for 0 < x < 1, and so on. Understanding this concept is crucial when evaluating the function at specific points or analyzing its behavior across its domain.

Limits of a Function

A limit can be thought of as the value that a function f(x) approaches as 'x' approaches a particular point. It is central to calculus and helps in understanding the behavior of functions at points where they might not be directly defined or may behave erratically.

In our step-by-step solution, we computed the limit of the piecewise function f(x) as x approaches 1. It's important to observe that limits can exist even if the function doesn’t have an actual value at that point - they represent what the function is approaching, not necessarily where it is. In the solution, it was determined that \( \lim _{x \rightarrow 1} f(x) = 2 \) by checking both sides of 'x' equal to 1 and ensuring they converge to the same value.

Continuity of a Function

Continuity is a smooth, uninterrupted path in a function without any holes, jumps, or vertical asymptotes at a point. For a function to be continuous at a particular point, three criteria must be met: the function must be defined at that point, the limit as 'x' approaches the point must exist, and the function’s value at that point must equal the limit.

Applying these criteria to the function f(x) from our exercise, we examined if it is continuous at x = 1. Though f(x) is defined at x = 1 and the limit as 'x' approaches 1 exists, the value of f(1) does not match the computed limit. Therefore, f(x) fails the continuity test at this point, indicating a disruption in the function's path.

Limit at a Point

When evaluating the limit at a point, we're not just looking at what value a function takes at that specific 'x', sometimes it’s more about the journey than the destination. The limit tells us where the function is headed as 'x' gets infinitely close to a certain value from both directions (approaching from the left and right).

In our exercise, we looked at the limit of f(x) as 'x' approaches 1. We found the left-hand limit (2*1 = 2) and right-hand limit (-2*1 + 4 = 2) both lead to the same value, indicating that the overall limit at that point is indeed 2. This emphasizes how limits capture the approaching value and are essential to understanding functions’ behaviors near points of interest.