From 11:00PM PDT on Friday, July 1 until 5:00AM PDT on Saturday, July 2, the Shmoop engineering elves will be making tweaks and improvements to the site. That means Shmoop will be unavailable for use during that time. Thanks for your patience!
is continuous at each given value. If not, explain.
x = -5
x = -4
x = 0
x = 2
x = 3
When we try to evaluate f(-5), we find
which is undefined. Since f(-5) does not exist, f is discontinuous at x = -5.
When we try to evaluate f(-4), we find f(-4) = 0, which is fine. We'll see if exists and agrees with f(-4). As x approaches -4 from the left, we use the part of the function definition that says The left-sided limit is
When x approaches -4 from the right, x is just a bit bigger than -4, so we'll use the part of the function definition that says f(x) = x – 8 for -4 < x ≤ 2. Then the right-sided limit is
Since the left- and right-hand limits agree,
Alas, since f(-4) = 0, the limit disagrees with the function value, therefore f is discontinuous at x = -4.
f(0) = 0 – 8 = -8. As x approaches 0 from the left or the right, we use the part of the function definition that says f(x) = x – 8. Since the function is continuous at x = 0.
f(2) = 2 – 8 = -6, which is fine. Now we'll check the limit. As x approaches 2 from the left we use the part of the function's definition that says f(x) = x – 8 for -4 < x ≤ 2, and find
As x approaches 2 from the right we use the part of the function definition that says
The bottom of this expression factors as (x – 2)(x – 3), and from earlier work with limits we know that is undefined. Therefore does not exist, and f is discontinuous at x = 2.
When we try to evaluate f(3), we find which is undefined. Therefore f is discontinuous at x = 3.
For what values of x is the function discontinuous.
f is defined everywhere, so there can't be any values of x where f is discontinuous from being undefined. That means we need to look for places where the limit of f is undefined or disagrees with the value of f. The only place this could possibly happen is x = 0. Let's investigate. The left-hand limit is
The right-hand limit is
The one-sided limits agree,
Alas, this disagrees with the function value, f(0) = 0. The function is discontinuous at x = 0.
For the function, determine all values at which the function is discontinuous.
First, we'll look for values any of x where g is undefined. We're fine at x = 1, since g(1) = 2 by the second piece of the definition, but x = 2 is not fine. The function definition says what should happen for x < 2 and 2 < x, but doesn't make any provision for what happens at x = 2. Therefore g is undefined, and therefore discontinuous, at x = 2.
Now we will look for places where the limit is undefined or disagreeable. Since is undefined, is also undefined and g is discontinuous when x = 1. That's the only troublesome spot we need to check.
To summarize, the function g is discontinuous at x = 1 and x = 2.
For what values of x is h(x) discontinuous.
The function definition says
The denominator factors, therefore we could rewrite this as
This function is undefined, and therefore discontinuous, at x = 4 and x = 5.The only place limits could go wrong, besides x = 4 and x = 5, is when x = 1. At x = 1 we have
As x approaches 1 from the right we have
Since the function isn't defined for values less than 1, we can't actually take a left-sided limit. In cases like this, we just use the right-sided limit as the value of the limit from all directions.
This agrees with h(1) = 1,
so the function is continuous at x = 1.
To summarize, h is only discontinuous at x = 4 and x = 5.