# function notation question

#### bumblebee123

##### Junior Member
can anyone help to explain this question?

question: f(x) = x^2 + 2x + 3, x ≥ 0

a) write x^2 + 2x + 3 in the form (x + c)^2 + d

I managed to do this: ( x + 1 )^2 + 2

b) the domain of f is x ≥ 0 Find the range of F

but this is where I get stuck. Can anyone help explain what to do next?

#### MarkFL

##### Super Moderator
Staff member
One way we could proceed is to begin with the given domain:

$$\displaystyle 0\le x$$

$$\displaystyle 1\le x+1$$

Since both sides are positive, we may square them

$$\displaystyle 1\le(x+1)^2$$

$$\displaystyle 3\le(x+1)^2+2=f(x)$$

#### Dr.Peterson

##### Elite Member
can anyone help to explain this question?

question: f(x) = x^2 + 2x + 3, x ≥ 0

a) write x^2 + 2x + 3 in the form (x + c)^2 + d

I managed to do this: ( x + 1 )^2 + 2

b) the domain of f is x ≥ 0 Find the range of F

but this is where I get stuck. Can anyone help explain what to do next?
Another way is to graph (or just imagine graphing) the function. From the rewritten form (sometimes called "vertex form"), you can see that the minimum value for the unrestricted function is 2, attained at x=-1. The function increases for all x ≥ -1, which implies that it is increasing on the stated domain, x ≥ 0, and must attain its minimum value at the left-hand end of that domain, namely at x = 0. So the minimum value is f(0) = 3.

You could also use calculus to determine all the same facts and make the same conclusion.

#### bumblebee123

##### Junior Member
Another way is to graph (or just imagine graphing) the function. From the rewritten form (sometimes called "vertex form"), you can see that the minimum value for the unrestricted function is 2, attained at x=-1. The function increases for all x ≥ -1, which implies that it is increasing on the stated domain, x ≥ 0, and must attain its minimum value at the left-hand end of that domain, namely at x = 0. So the minimum value is f(0) = 3.

You could also use calculus to determine all the same facts and make the same conclusion.
how did you know that the minimum value for the function is 2, attained at x= -1?

is it because ( -1 + 1)^2 + 2 = 2

#### bumblebee123

##### Junior Member
here's what I've worked out so far:

it tells me that the domain of f is x≥ 0

to find x, if i use the completed square, x + 1 = 0
x = -1

if i put the x coordinate back into the equation: ( -1 + 1 )^2 +2 = 2

but this is when x = -1, so when x = 0 I need to +1 to the answer which = 3

so the range is f(x) ≥ 3

I also found another method I tried gives me the same answer, is this a correct method? :

as the domain is x ≥ 0

if I put x = 0 into the completed square equation: ( 0 + 1 )^2 + 2 = 3

#### HallsofIvy

##### Elite Member
Yes, $$\displaystyle (x+ 1)^2+ 3$$ is smallest when $$\displaystyle (x+ 1)^2$$ which is when $$\displaystyle x+ 1$$ is smallest which is when x= 0. (Notice that if the condition "$$\displaystyle x\ge 0$$" where not there, $$\displaystyle (x+ 1)^2$$ would be smallest when x=-1.)

#### Dr.Peterson

##### Elite Member
how did you know that the minimum value for the function is 2, attained at x= -1?

is it because ( -1 + 1)^2 + 2 = 2
The way I suggested there depends on familiarity with "vertex form", $$\displaystyle y = (x - h)^2 + k$$. From that form, you can read off that the vertex is $$\displaystyle (h, k)$$.

If you don't know that, then you would what has been discussed on your other thread about quadratic functions. We know that the smallest value of $$\displaystyle (x + 1)^2$$ occurs when $$\displaystyle x + 1$$ is zero (because otherwise it is always positive). Then, as you showed, "plugging in" $$\displaystyle x = -1$$ gives the corresponding value of y.

#### HallsofIvy

##### Elite Member
A real number, squared, is never negative. Whatever x and h are, $$\displaystyle (x- h)^2\ge 0$$, equal to 0 when x= h and positive otherwise. So whatever x and h are, $$\displaystyle (x- h)^2+ k\ge k$$, equal to k when x= h and greater than k otherwise.

#### bumblebee123

##### Junior Member
The way I suggested there depends on familiarity with "vertex form", $$\displaystyle y = (x - h)^2 + k$$. From that form, you can read off that the vertex is $$\displaystyle (h, k)$$.

If you don't know that, then you would what has been discussed on your other thread about quadratic functions. We know that the smallest value of $$\displaystyle (x + 1)^2$$ occurs when $$\displaystyle x + 1$$ is zero (because otherwise it is always positive). Then, as you showed, "plugging in" $$\displaystyle x = -1$$ gives the corresponding value of y.
so I would've known either by because k in the equation = 2

or because plugging in x=-1 gives y= 2 ?

#### Dr.Peterson

##### Elite Member
so I would've known either by because k in the equation = 2

or because plugging in x=-1 gives y= 2 ?
Yes, on both counts.