A relation is (also) a function if every input x is mapped to a unique output y.
In terms of graphical representation, this implies that a graph represents a function if there doesn't exist a vertical line that intersects the graph more than once. So:
- The first graph is exactly a vertical line, so it's not a function.
- The second graph represents the function y=x, so it's a function: you can see that every possible vertical line crosses the graph only once.
- The third graph is not a function, because you can draw vertical lines that cross the graph twice.
- Similarly, in the fourth graph you can draw vertical lines that cross the graph twice
- The fifth graph is a function, because every vertical line crosses the graph once
- The last graph is a function, although discontinuous, for the same reason.
Answer:
No.
Step-by-step explanation:
If something is <em>less</em> than 5, no, however if something is less than or equal to (≤ sign), then yes.
It’s written as 0.6 in decimal form
Answer:
![(f o g)(x) = 3x + \frac{2}{3}](https://tex.z-dn.net/?f=%28f%20o%20g%29%28x%29%20%3D%203x%20%2B%20%5Cfrac%7B2%7D%7B3%7D)
Step-by-step explanation:
We have the function
and we have the function
. We want to find g(x) composed with f(x)
Then, the function (f o g)(x) is the same since f(g(x))
That is, you must do x = g(x) and then enter g(x) into the function f(x).
![f(g(x)) = (g(x)) -\frac{1}{3}](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%28g%28x%29%29%20-%5Cfrac%7B1%7D%7B3%7D)
![f(g(x)) = (3x + 1) -\frac{1}{3}](https://tex.z-dn.net/?f=f%28g%28x%29%29%20%3D%20%283x%20%2B%201%29%20-%5Cfrac%7B1%7D%7B3%7D)
Simplifying, we obtain:
![(f o g)(x) = 3x + 1 -\frac{1}{3}\\\\(f o g)(x) = 3x + \frac{2}{3}](https://tex.z-dn.net/?f=%28f%20o%20g%29%28x%29%20%3D%203x%20%2B%201%20-%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C%28f%20o%20g%29%28x%29%20%3D%203x%20%2B%20%5Cfrac%7B2%7D%7B3%7D)
Finally. The composite function is:
![(f o g)(x) = 3x + \frac{2}{3}](https://tex.z-dn.net/?f=%28f%20o%20g%29%28x%29%20%3D%203x%20%2B%20%5Cfrac%7B2%7D%7B3%7D)