Use the function f defined and graphed below to answer the...

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 $x=-1 ?$

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 $x=-1 ?$

Key Concepts

Piecewise-Defined Functions

A piecewise-defined function is defined by different expressions over different parts of its domain. This concept requires checking which part of the definition applies when evaluating the function at a specific point and can lead to special considerations at the boundaries between pieces.

One-Sided Limits

One-sided limits involve evaluating the behavior of a function as the input approaches a specific point from one direction (either from the left or the right). This is important for functions defined on limited subdomains or at endpoints, where the full two-sided limit might not exist.

Function Value at a Point

The function value at a point is the output obtained by substituting the point into the function's rule, if the point is included in the domain. This concept is essential in determining the behavior of the function at a specific input, especially for piecewise-defined functions where the definition can change at particular points.

Continuity at a Point

Continuity at a point requires that the function's value at the point and the function's limit as the input approaches the point (from appropriate directions) are equal. This concept is fundamental in calculus because it defines a smooth connection between values of a function, ensuring that there are no jumps or gaps at that point.

Transcript

00:01 So we're going to consider a lot of different properties within the given function.

00:04 First, we want to know if f of negative 1 exists.

00:07 So looking at the functions given to us, we see that yes, it does exist because it's going to be, with regards to the first problem or the first equation, we see it holds over the interval where negative 1 is less than or equal to x.

00:22 So it does exist.

00:24 The limit as x approaches from the right, as x approaches negative 1 from the right exists.

00:29 And we see it is going to be the same as f of negative 1.

00:35 But we don't say that f is continuous because we don't know the limit as x approaches negative 1 from the left.

00:46 And that's what we have to look out for.

00:49 Then we consider f of 1.

00:53 We see that f of 1 does exist because we see that f of 1 equals 1.

01:00 We say that the limit does not exist though because as we approach actually the limit will exist the limit will be two however we see that this is not continuous because the function value is one but the limit value is two so since the limit doesn't equal the function value it is not continuous and then um we see the function is not defined at two because if we look at the intervals x is never equal to 2 so it can't possibly be continuous at 2.

01:33 We see the function is continuous on all values except for all of values between negative 1 and 3, with the exception of 0, 0, 1, 2, and 3.

01:56 So all of those integer values, it's not going to be continuous.

02:02 And then lastly, we want to know what value should be assigned to f of 2 to make the function continuous...

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