y = x + 4/x
replace x with -x. Do you get back the original equation after simplifying. if you do, the function is even.
replace y with -y AND x with -x. Do you get back the original equation after simplifying. If you do, the function is odd.
A function can be either even or odd but not both. Or it can be neither one.
Let's first replace x with -x
y = -x + 4/-x = -x - 4/x = -(x + 4/x)
we see that this function is not the same because the original function has been multiplied by -1
. Let's replace y with -y and x with -x
-y = -x + 4/-x
-y = -x - 4/x
-y = -(x + 4/x)
y = x + 4/x
This is the original equation so the function is odd.
Answer:
If that's the whole question then what exactly are we trying to solve here?
Step-by-step explanation:
<u>Corrected Question</u>
Is the function given by:
continuous at x=4? Why or why not? Choose the correct answer below.
Answer:
(D) Yes, f(x) is continuous at x = 4 because 
Step-by-step explanation:
Given the function:

A function to be continuous at some value c in its domain if the following condition holds:
- f(c) exists and is defined.
exists.
At x=4
Therefore: 
By the above, the function satisfies the condition for continuity.
The correct option is D.
<h2>Answer</h2>

Or as ordered pairs: 
<h2>Explanation</h2>
Lets solve our system of equations step by step
equation (1)
equation (2)
1. Solve for
in equation (2)

equation (3)
2. Replace equation (3) in equation (1) and solve for 




or
3. Replace the values of
in equation (3) and solve for 
- For 


or 
- For 



or 
So, the solutions of our system of equation are:

Detecting...<span>
</span>Your answer would be: