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Range of a function

  • #1
Help needed.

Homework Statement


Functions g and h are defined as follows:
g : x → 1 + x x ∈ R
h : x → x² + 2x x ∈ R

Find i.) the ranges of g and h,

ii.) the composite functions h ° g and g ° h, stating their ranges.
Not sure how this is to be done help needed, please.

2. The attempt at a solution

i.) range of g => R = {y : y ∈ R}

1 + x = 0
x = -1
1 - 2 = -1
-b/2a = -2/2 = -1
range of h => R = {y : y ≥ - 1, y ∈ R}
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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The way to do this explicitly, would be to simply give the formula for h o g and g o h, which will give you two quadratic functions again.

But you can also find it by argument. For example, consider h o g. You take some x and apply g. What are the possible values y that you get. Then apply h. What can y map to?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Help needed.

Homework Statement


Functions g and h are defined as follows:
g : x → 1 + x x ∈ R
h : x → x² + 2x x ∈ R

Find i.) the ranges of g and h,

ii.) the composite functions h ° g and g ° h, stating their ranges.
Not sure how this is to be done help needed, please.

2. The attempt at a solution

i.) range of g => R = {y : y ∈ R}

1 + x = 0
x = -1
1 - 2 = -1
-b/2a = -2/2 = -1
range of h => R = {y : y ≥ - 1, y ∈ R}
h(x)= [itex]x^2+ 2x= x^2+ 2x+ 1- 1= (x+1)^2- 1[/itex]
I presume that is why you were looking at "1+ x= 0"!

Now, g(x) can be any number so h(g(x)) can be what?

h(x) must be larger than or equal to 1 so g(h(x)) can be what?
 
  • #4
Not sure how much of a difference this is:
g : x |→ 1 + x x ∈ R
h : x |→ x² + 2x x ∈ R

This would be the range of h(x)=> [itex] x^2 + 2x = x^2 + 2x + 1 - 1 = (x + 1)^2 - 1[/itex]
and my range for g is correct?

Now, g(x) can be any number so h(g(x)) can be what?
h(g(x)) can be any number. R = {y: x ∈ R}

h(x) must be larger than or equal to 1 so g(h(x)) can be what?
g(h(x)) can be larger than or equal to 1. R = {y : ≥ - 1, y ∈ R}
 
Last edited:
  • #5
CompuChip
Science Advisor
Homework Helper
4,302
47
Your result for h(x) already was correct (it's {y | y ≥ -1}).
Halls was just pointing out that h(x) = g(x)2 - 1, I suppose (which you could also have used to obtain the same result).

Note that h(g(x)) means you are first evaluating g on x. This can give you any number, which you plug into h...
 

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