Starting with Spivak - question

[robjow]

Hey, I'm just starting out with calculus and am giving the Spivak book a try per threads on here. In the introduction to functions chapter spivak defines a function f(x) =x^2. for all x such that -17 ≤ x ≤ π/3
Then says you should be able to check the following assertion about the function defined above:

f(x+1) = f(x) + 2x + 1 if -17 ≤ x ≤ (π/3) - 1

I don't understand why there a negative one is now needed at the end of the domain

 

[lurflurf]

-18 ≤ x ≤ (π/3) - 1 is the domain of f(x+1)
-17 ≤ x ≤ (π/3) is the domain of f(x)
-17 ≤ x ≤ (π/3) - 1 is the domain the two share

edit:to correct error

 

[robjow]

thanks for your reply,

That doesn't make sense to me.

My incorrect thinking goes like this..

What is f(x+1) actually saying? To me its saying: apply the function f (which is to square) to the domain x (defined above) +1..making the domain of f(x+1) -16 to π/3 + 1

 

[robjow]

Shouldn't the domain of f(x+1) be -18 ≤ x ≤ (π/3) - 1??

 

[Mark44]

That doesn't make sense to me.

My incorrect thinking goes like this..

What is f(x+1) actually saying? To me its saying: apply the function f (which is to square) to the domain x (defined above) +1..making the domain of f(x+1) -16 to π/3 + 1

The graph y = f(x + 1) is the graph of y = f(x) shifted (translated) one unit to the left. Since the domain of y = f(x) is -17 ≤ x ≤ π/3, the domain of y = f(x + 1) will be -18 ≤ x ≤ π/3 - 1. IOW, the original interval shifted one unit to the left.

Forget the weird domain for a moment. You know what the graph of y = f(x) = x² looks like, right? The graph of y = g(x) = f(x + 1) is a parabola the opens up, and whose vertex is at (-1, 0). Every point on the shifted parabola is one unit to the left of its corresponding point on the graph of y = f(x).

Shouldn't the domain of f(x+1) be -18 ≤ x ≤ (π/3) - 1??

Yes, which is different from what you said above.