Answer:
ok thx
Step-by-step explanation:
Answer:
Do you want to be extremely boring?
Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?
is a valid solution.
Want something more fun? Why not a parabola?
.
At this point you have three parameters to play with, and from the fact that
we can already fix one of them, in particular
. At this point I would recommend picking an easy value for one of the two, let's say
(or even
, it will just flip everything upside down) and find out b accordingly:![f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1](https://tex.z-dn.net/?f=f%281%29%3D2%20%5Crightarrow%201%5E2%2Bb%2B2%3D2%20%5Crightarrow%20b%3D-1)
Our function becomes
Notice that it works even by switching sign in the first two terms: ![f(x) = -x^2+x+2](https://tex.z-dn.net/?f=f%28x%29%20%3D%20-x%5E2%2Bx%2B2)
Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: ![f(x) = A cos (kx)](https://tex.z-dn.net/?f=f%28x%29%20%3D%20A%20cos%20%28kx%29)
Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need
, and at that point the first condition is guaranteed; using the second to find k we get ![2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi](https://tex.z-dn.net/?f=2%3D%202%20cos%20%28k1%29%20%3D%20cos%20k%20%3D%201%20%5Crightarrow%20k%20%3D%202%5Cpi)
![f(x) = 2cos(2\pi x)](https://tex.z-dn.net/?f=f%28x%29%20%3D%202cos%282%5Cpi%20x%29)
Or how about a sine wave that oscillates around 2? with a similar reasoning you get
![f(x)= 2+sin(2\pi x)](https://tex.z-dn.net/?f=f%28x%29%3D%202%2Bsin%282%5Cpi%20x%29)
Sky is the limit.
The answer to the math question
Answer:
2/1 or 2
Step-by-step explanation:
8/4 is 2
9/9 is 1