Confused about: g(x) = (x - 1)^2

[Googl]

I am revising on sketching curves. But I am very confused about the following even though I have become very good at sketching graphs.

An example from the book is:

Given that:
i) f(x) = x^3
ii) g(x) = x(x - 2)

Sketch the curves with equation y=f(x + 1) and g(x +1)

I don't understand the example that the book provides. Not because I don't know how to sketch the curve but the use of g(x) is what I don't understand. Somewhere in the example workout it says:

g(x) = (x - 1)^2
so -> g(x) = f(x - 2)

Also
h(x) = x^2 + 2
so -> h(x) = f(x) + 2

Please help as I would like to finish this chapter sooner. The use of h(x) and g(x) is what I don't understand. I will also appreciate any links that you may provide.

Thanks.

 

[Fredrik]

I am revising on sketching curves. But I am very confused about the following even though I have become very good at sketching graphs.

An example from the book is:

Given that:
i) f(x) = x^3
ii) g(x) = x(x - 2)

Sketch the curves with equation y=f(x + 1) and g(x +1)

I don't understand the example that the book provides. Not because I don't know how to sketch the curve but the use of g(x) is what I don't understand.

I assume that "g(x+1)" really means y=g(x+1). Just use the definition of g (given in ii) to rewrite the right-hand side.

g(x) = (x - 1)^2
so -> g(x) = f(x - 2)

There's something missing here. This is clearly not the same g as before, so it's probably a different f as well. I can't comment unless you tell me what f is.

Also
h(x) = x^2 + 2
so -> h(x) = f(x) + 2

Same thing here. What is f?

Edit: Some general comments about functions: Students often think of functions as relationships between variables, but it's better to think of a function as a rule that associates a member of a set (called the function's codomain) with each member of a set (called the function's domain). If f is a function, f(x) denotes the member of the codomain that f associates with x. f(x) is called the value of f at x. We also say that f takes x to f(x).

For example, if f is defined by $f(x)=x^2$ for all real numbers x, then f(3)=9. Here f is the function that takes every number to its square, and its value at 3 is 9. Also, the value of f at x+1 is f(x+1), which by definition of f is =(x+1)². Note that f(x+1) isn't a function. It's a member of the codomain of f. However, without knowledge of the value of x, we can't know which one it is.

We can also talk about the function that takes x to f(x+1). If we define g(x)=f(x+1) for all x such that x+1 is in the domain of f, then we have defined a function g. People often refer to that function as "f(x+1)", but if you want to be accurate, you need to refer to it as "the function that takes x to f(x+1)", or something like that.

 

[Googl]

Thanks for the reply. It is beginning to make sense now.

 

[Square1]

Hey. This is the first time I've ever helped someone on this site. I'll try my best.

Given that:
i) f(x) = x^3
ii) g(x) = x(x - 2)

Sketch the curves with equation y=f(x + 1) and g(x +1)

So is the problem here trying to figure out what these two new f's and g's are supposed to be?

When you have f(x) = (x)^3, think of the left hand side as saying you are making a rule labelled f on the variable also called your input, or argument, (x). This is denoted f(x). The right hand side says specifically what that rule involving your input is. (x)^3.

By the way, we could have rules that do a lot more things, like cubing x, then adding another 6x to it, and perhaps adding a constant to it all like 5, then perhaps dividing everything by 2. This would look like f(x) = (x^3 + 6x + 5)/2 . Try to understand this maybe after if it doesn't make sense now.

Continuing on though...
y = f(x+1) is saying take your original rule that you were given on (x) ( the rule is to cube (x) right? ), and just do the same rule to a new the input, (x+1), and it also is saying that were calling this new thing y. So in this case we just slap on ^3 to the input (x+1).

I'm not really sure what you mean by the other functions they are giving you either.

Hope this helped.