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### Course: Differential Calculus>Unit 1

Lesson 12: Continuity over an interval

# Functions continuous at specific x-values

Determine the continuity of two functions, ln(x-3) and e^(x-3), at x=3. Explore the concept of continuity, highlighting that common functions are continuous within their domain. Discover that ln(x-3) is not continuous at x=3, while e^(x-3) is continuous for all real numbers, including x=3.

## Want to join the conversation?

• Im sure this is a silly question and Im missing the obvious, but why did you say g(3)=ln(0)? How did you know to put zero? Are you saying that nothing raised to a power equals three therefore its zero? Thank you!
• Because g(x)=ln(x-3) so when x=3, the function equals g(3)=ln(3-3)=ln(0).
• what is meant by the natural logarithm?
• On the practice for this section, it asks us to know if a function is continuous for all real numbers.. one of the options is

cubic root of (x+1)

And it is at f(x) at -2... how can you take a cubic root of -1? isn't that an imaginary value? and thus not real?
• Just as square root is the inverse of squaring, cube root is the inverse of cubing.
Cube root of -1 is -1, because (-1)^3 = -1 * -1 * -1 = -1

You can't have a square root of a negative number, this would result in imaginary number. This is true and extends to all even roots, i.e. square root, 4th root, 6th root, so on and so on. But imaginary number only applies to even roots. You can have cube root or any odd roots of a negative number. Cube root of -1 is one example. Cube root of -8 is -2, because -2 * -2 * -2 = -8.

Hope that helps.
• What if a specific x-value is defined but the point still not continuous? Obvoisly, you could tell that the point is not continous by graphing the function, but is there a way to tell just by looking at the equation? Also, is there an example of an equation where a point is defined but not continuous?
• An example of a function where a point is defined but is not continuous is the Heaviside step function.
• In the Practise, I found this hint for a question that doesn't make sense to me. It says: f(x)=tan(x)=sin(x)/cos(x) is defined for all numbers except when cos(x) is 0. It makes sense-you can't divide by zero but then it says: so it's domain is all x-values such that x does not equal *(2k+1)Xpi/2* where k is an integer.

• So, you understand the basic idea. f(x)=tan(x) is not defined when cos(x) =0, because tan(x) is sin(x)/cos(x).

Let's look at where cos(x) =0. You'll need to draw the graph of cos(x) for yourself. You'll see that in a 2pi interval, there are two points where cos(x) = 0. Those points are pi/2 and 3pi/2.

Now, if you were to extend the graph further, you'd see that 5pi/2, 7pi/2 are also zeros.

So, let's write down a sequence which defines where cos(x) =0:
pi/2, 3pi/2, 5pi/2, 7pi/2...

Notice anything about the sequence? They're odd numerators times pi with denominator 2.

So, the formula then becomes:
(2k+1)Pi/2 where k is any integer
• Is it fair to say that while the function f(x) = √x is continuous on the interval [0,∞), it is not continuous at the point x = 0 because the two-sided limit doesn't exist?
• For a function to be continuous at the endpoints of its domain it's enough for the one-sided limit to equal the function value.
• At x = 3 he says it is undefined and draws an asymptote how does the function cross a vertical asymptote? I thought functions can only cross horizontal and slant asymptotes and can never cross vertical ones
• The asymptote is for g(x) while the function f(x) crosses the asymptote. g(x) approaches the asymptote from the right side but never touches it. In fact, it does not matter whether asymptote is vertical, slant, or horizontal. No function will cross its own asymptote.
• At 4.09, the graph is shown extending into negative values. Is this correct? As x takes on higher negative values, y values asymptotically approach zero, isn't it? In fact there is no way a negative value can be found for function f(x)
• All exponential functions have a range f(x) > 0. Not only e^x or e^(3-x), but 2^x, 10^x, etc., will have outputs everywhere greater than zero.

Can you explain why this is true?
• How will we identify whether a function is defined or not?
(1 vote)
• A function itself cannot be defined/undefined. However, it can be defined/undefined *at a point* (the terminology is important).

To check for whether it is defined or not, simply plug that point into the function. For example, if I have $f(x) = \frac{1}{x}$ and x = 0, I get $f(0) = \frac{1}{0}$. Clearly, $\frac{1}{0}$ isn't defined. So, the function isn't defined at x = 0, which is clearly shown by the graph as it asymptotes towards infinity as x gets closer to 0.