ihatePE
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#1
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repost cos the other one is a mess

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how would you find out the range?

i have the markscheme so i dont need the right answer, i just want to know how they arrived there because the step wasnt in the MS.
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ihatePE
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this is how i would usually do it (based on the many questions ive done)

the Domain of fg is the domain of g(x) therefore D(fg) (0, infinity)
the range of fg is the range of g(x) R(-3, infinity) applied to f(x) so f(x)= (-3)^2 -25 = -16
and f(x) =(infinity)^2 - 25 = infinity so

R(fg) = (-16, infinity)
which isnt right accoridng to the mark scheme
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Notnek
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(Original post by ihatePE)
this is how i would usually do it (based on the many questions ive done)

the Domain of fg is the domain of g(x) therefore D(fg) (0, infinity)
the range of fg is the range of g(x) R(-3, infinity) applied to f(x) so f(x)= (-3)^2 -25 = -16
and f(x) =(infinity)^2 - 25 = infinity so

R(fg) = (-16, infinity)
which isnt right accoridng to the mark scheme
You have the range of g(x) correct.

So you're looking for the range of f(x)=x^2-25 for domain (-3,\infty).

For a lot of functions, you can't just plug in the max/min of the domain and expect to get the range - you need to actually think what the biggest and smallest values of f(x) are if the domain is (-3,\infty).

So for example, you say the range is (-16,\infty) but if I plug in -2 (which is within the domain) into f(x) I get f(-2) = -21 and this is less than -16. So the range can't be (-16,\infty).

Have a go at correcting this. It may help if you sketch a graph of f(x).

Also, the domain of fg is not always the domain of g. That can be explained after you understand the range.
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ihatePE
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(Original post by Notnek)
You have the range of g(x) correct.

So you're looking for the range of f(x)=x^2-25 for domain (-3,\infty).

For a lot of functions, you can't just plug in the max/min of the domain and expect to get the range - you need to actually think what the biggest and smallest values of f(x) are if the domain is (-3,\infty).

So for example, you say the range is (-16,\infty) but if I plug in -2 (which is within the domain) into f(x) I get f(-2) = -21 and this is less than -16. So the range can't be (-16,\infty).

Have a go at correcting this. It may help if you sketch a graph of f(x).

Also, the domain of fg is not always the domain of g. That can be explained after you understand the range.
I totally forgot about this problem with quadraticss!!!! i usually always draw graphs but when it comes to function it almost always escape my mind, thanks
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Notnek
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(Original post by ihatePE)
I totally forgot about this problem with quadraticss!!!! i usually always draw graphs but when it comes to function it almost always escape my mind, thanks
You're not the only one - this is a common mistake It's always best to think about the range and not just plug in the mix/max of the domain. Drawing a sketch is often useful if you're unsure.

By the way, here's an example where the domain of fg is not the same as the domain of g.

g(x) = x+3,  \ \ x\in \mathbb{R}

f(x) = \sqrt{x}, \ \ x>0

You would say that the domain of fg(x) = \sqrt{x+3} is x\in \mathbb{R} but can you see why this can't be correct?
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ihatePE
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(Original post by Notnek)
You're not the only one - this is a common mistake It's always best to think about the range and not just plug in the mix/max of the domain. Drawing a sketch is often useful if you're unsure.

By the way, here's an example where the domain of fg is not the same as the domain of g.

g(x) = x+3,  \ \ x\in \mathbb{R}

f(x) = \sqrt{x}, \ \ x>0

You would say that the domain of fg(x) = \sqrt{x+3} is x\in \mathbb{R} but can you see why this can't be correct?
mhmmmm, it's a root, therefore only positive values can be in it :hoppy:
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Notnek
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(Original post by ihatePE)
mhmmmm, it's a root, therefore only positive values can be in it :hoppy:
Exactly. So the domain of fg is the set of values that are in the domain of g and are mapped to something by fg. The domain of fg is always a subset of the domain of g and may be equal to it depending on the question.
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