Question

Please help! Given h(x)=(x-2)^2, find the domain of the function. The function is increasing for?

Answer 1

Is there any point x where the function will not be known / not exist?

Answer 2

The domain is all real numbers, correct?

Answer 3

Yes. There isn't a point where the function disappears -- that usually only happens when we have variables in the denominator or even-index roots with variables.

You need to know where the function is increasing?

Answer 4

Oh o.k. :)
Yes, that's what I need to know

Answer 5

All x?

Answer 6

No, this is a parabola. It has points where it is both increasing and decreasing.

Hmm.. you are doing this in Calculus? there's a way to go about it without using calc, but you may be more familiar with a calculus method. :P

Answer 7

:/ Oh, wait. Here are options:

x<2

all x

no x

x>2

Answer 8

If we take the derivative of h(x), we'd get...
h'(x) = 2(x-2)^1 * 1
= 2x - 4.

The function is increasing when we have a positive derivative. So...
2x - 4 > 0
2x > 4
x > 2

The other method I was thinking about was just finding the vertex since its in vertex form already (Vertex is V(2,0)). This is a minimum value since the parabola points upwards. So, the values to the right of x (x>2; 3, 4... etc) are where the function starts to become more positive.

Answer 9

Oh, I see. :) Thanks so much for the explanations AccessDenied!

Answer 10

No problem. I hope it makes sense. :)

Answer 11

One more question: So it'd be decreasing for x<2?

Answer 12

Yes.

Answer 13

Thanks :)