Question 1

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!

Accepted Answer

Because g(x)=ln(x-3) so when x=3, the function equals g(3)=ln(3-3)=ln(0).

Question 2

what is meant by the natural logarithm?

Accepted Answer

It's the logarithm whose base is Euler's number e. The natural log of x is denoted ln(x).

Here's the video: https://www.khanacademy.org/math/algebra2/exponential-and-logarithmic-functions/e-and-the-natural-logarithm/v/natural-logarithm-with-a-calculator

Question 3

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?

Accepted Answer

An example of a function where a point is defined but is not continuous is the Heaviside step function.

Question 4

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?

Accepted Answer

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.

Question 5

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.

How do u get *(2k+1)Xpi/2*? Please help!

How do u get *(2k+1)Xpi/2*? Please help!

Accepted Answer

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

Question 6

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?

Accepted Answer

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.

Question 7

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

Accepted Answer

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.

Question 8

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)

Accepted Answer

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?

Question 9

How will we identify whether a function is defined or not?

Accepted Answer

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.

Question 10

wait why would f(x)=e^(x-3) be continuous at x=3? because f(3)=e^0, which is just 1. but the limit is approaching 3, so it doesn't meet the requirements of lim f(x) as x->a = f(a)? sorry if i'm missing something really obvious but i'm confused-- graphically i can understand it but it doesn't work for the equation

Accepted Answer

To determine if a function is continuous at a point, we need to ensure three conditions are met:

`1. The function is defined at that point.`
`2. The limit of the function as x approaches that point exists.`
`3. The limit as x approaches that point is equal to the value of 
 the function at that point.`

Let us check:
1. Is the function defined at x = 3? We can find out by substituting 3 for x.
`f(3) = eˣ⁻³ = e³⁻³ = e⁰ = 1 ∴ f(3) exists and equals 1.`

2. Does the limit of the function as x approaches 3 exist?
`limₓ→₃ eˣ⁻³ = e³⁻³ = e⁰ = 1 ∴ the limit exists and equals 1`

3. Does the limit as x approaches 3 equal the value of the function at that point?
`From 1 and 2, we have shown that f(3) and limₓ→₃ both are equal to 1`

Because these three conditions are met, f(x) = eˣ⁻³ is continuous at x = 3.

Course: AP®︎ Calculus AB (2017 edition) > Unit 1

Functions continuous at specific x-values

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