The function is
![f(x) = \begin{cases} x^2-4 &\mbox{if } x\ \textless \ 1, \\ x+4 & \mbox{if } x\ \textgreater \ 1. \end{cases}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cbegin%7Bcases%7D%20x%5E2-4%20%26%5Cmbox%7Bif%20%7D%20x%5C%20%5Ctextless%20%5C%201%2C%20%5C%5C%20%0Ax%2B4%20%26%20%5Cmbox%7Bif%20%7D%20x%5C%20%5Ctextgreater%20%5C%201.%20%5Cend%7Bcases%7D)
.
To the left of 1 the function is a quadratic polynomial, to the right, it is a linear polynomial. Polynomial functions are always continuous, so the only candidate point for discontinuity is x=1.
The left limit is calculated with 1 substituted in
![x^2+4](https://tex.z-dn.net/?f=x%5E2%2B4)
, which gives 5.
The right limit, is computed using the rule for the right part of 1, that is x+4.
Thus, the right limit is 1+4=5.
So, both left and right limits are equal. Now if f(1) is 5, then the function is continuous at 1.
But the function is not defined for x=1, that is x=1 is not in the domain of the function. Thus, we have a "whole" (a discontinuity) in the graph of the function.
The reason is now clear:
Answer:<span> f(1) is not defined</span>
Total number = 6
Number of permutations = 6!
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
Answer:
B, C, and D
Step-by-step explanation:
First, we have to simplify 9 and 6 down to 3 and 2
After you just see which one works and which one doesn't.
Answer:
Step-by-step explanation:
<u>There is a few ways to go about doing this problem. I am going to setup a function.</u>
x = seconds
f(x) = distance = How far
f(x) = (3.0* 10^8) x
<u>Insert 10^5 where x is located</u>
(3.0* 10^8) x
<u>Multiply</u>
(3.0* 10^8) (10^5)
<u>Do the math</u>
(3.0* 10^8) (10^5) = 3 * 10^13