Am I Correctly Understanding the Function f(x) = x^2 + 1 for 0 < x < 6?

[Casio1]

f(x) = x^2 + 1 (0 < x < 6)

Please advise if I am understandng the above correctly!

f is the function of x

x is the domain

x^2 + 1 is the rule of the function

(0 < x < 6) these are the real numbers that satisfy x, so does that mean use as follows;

f(0) = 0^2 + 1 or

f(1) = 1^2 + 1 etc upto 6, or does it mean;

f(0) = 0^2 + 1 and then straight to

f(6) = 6^2 + 1

I hope you can follow what I mean. There is no point giving an example until I understand the basics.

 

[SuperSonic4]

f(x) = x^2 + 1 (0 < x < 6)

Please advise if I am understandng the above correctly!

f is the function of x

x is the domain

x^2 + 1 is the rule of the function

(0 < x < 6) these are the real numbers that satisfy x, so does that mean use as follows;

f(0) = 0^2 + 1 or

f(1) = 1^2 + 1 etc upto 6, or does it mean;

f(0) = 0^2 + 1 and then straight to

f(6) = 6^2 + 1

I hope you can follow what I mean. There is no point giving an example until I understand the basics.

$f$ is a convention meaning function (indeed you use g to denote a second function and so on)
$x$ is what the function is in terms of - this if often the same as the variable on the right - you can have f(t) for a function in t.
$x^2+1$ is the argument of the function - what we do to an input to generate a value
$(0 \leq x \leq 6)$ is the domain. This restricts possible values of the input - in this case between 0 and 6 inclusive

$f(0) = 0^2 + 1$ is your lower bound and $f(6) = 6^2+1$ is your upper bound. You can choose any number between 0 and 6 inclusive as an input. We can take $\pi$ as an example $(\pi \approx 3.14)$: $f(\pi) = \pi^2 +1$

 

[HallsofIvy]

f(x) = x^2 + 1 (0 < x < 6)

Please advise if I am understandng the above correctly!

f is the function of x

Okay, yes.

x is the domain

No, x is the "variable". The domain is the closed interval [0, 6], the set of all possible values for x. We know that because the definition of f says "0\le x\le 6".

x^2 + 1 is the rule of the function

Yes.

(0 < x < 6) these are the real numbers that satisfy x, so does that mean use as follows;

f(0) = 0^2 + 1 or

f(1) = 1^2 + 1 etc upto 6, or does it mean;

f(0) = 0^2 + 1 and then straight to

f(6) = 6^2 + 1

I hope you can follow what I mean. There is no point giving an example until I understand the basics.

I'm not sure what you mean by "straight up to". I think you mean that x can be any number between 0 and 6 rather than just integer values (which is what your first statements seem to mean). If that is what you mean then, yes, that is correct. For example 4.5 lies between 0 and 6 so is in the domain and f(4.5)= (4.5)^2+ 1= 20.25+ 1= 21.25.