Domain & Range help for my bestie (1 Viewer)

barnyard

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i did the same as you except i used the discriminant being -ive in x^2+2x+3 and then tested x=0 to find that x^2+2x+3 is +ive for all x values ie the domain is all real x
additionally, you can use the parabola's axis of symmetry, -b/2a, instead of using the derivative to find the max y value. taking the inverse will make the lowest y value the highest so the range will be from 0 non inclusive to 1/sqrt2 inclusive (i think that is ext1 graphing and idk if this is an adv question? all the maths gets confused in my head lol)
 

NotBamboo

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i did the same as you except i used the discriminant being -ive in x^2+2x+3 and then tested x=0 to find that x^2+2x+3 is +ive for all x values ie the domain is all real x
additionally, you can use the parabola's axis of symmetry, -b/2a, instead of using the derivative to find the max y value. taking the inverse will make the lowest y value the highest so the range will be from 0 non inclusive to 1/sqrt2 inclusive (i think that is ext1 graphing and idk if this is an adv question? all the maths gets confused in my head lol)
yes the -b/2a is very helpful
instead of all the messy derivative stuff
i didn't think of it cus it doesn't look like a parabola to me

also what do you mean inverse?
 

barnyard

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yes the -b/2a is very helpful
instead of all the messy derivative stuff
i didn't think of it cus it doesn't look like a parabola to me
well it was originally a parabola (in fact you even noted this in your working with your u sub) and now it's been transformed a couple times. the sqrt doesnt change anything about WHERE the lowest/highest y value is, just WHAT it is, and the inverse or whatever it is makes the lowest value become the highest and vice versa, leading to y approaching 0 at both x= +ive and -ive infinity
also what do you mean inverse?
1/f(x)
i forgor what it's called feel free to correct me lol
 
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liamkk112

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@Shavi Masee @jane1820
original question:

Find the domain and range for
View attachment 47397
well the domain is every real x, since x^2+2x+3 has no real zeroes so you can never divide by zero
the quadratic is also positive on it's domain, since at the vertex -2/2 = -1, the function is positive and it's a concave up parabola, so there's no restrictions on x

for the range, look at the function f(x) = x^2+2x+3. that has range [2,inf) since at the vertex f(-1) = 2 and f goes to infinity as x goes to +/- infinity.
hence 1/f will have range (0,1/2] since at +/- infinity 1/f is gonna go to zero. taking square roots then just makes the range (0, 1/sqrt(2)]
 

NotBamboo

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well the domain is every real x, since x^2+2x+3 has no real zeroes so you can never divide by zero
the quadratic is also positive on it's domain, since at the vertex -2/2 = -1, the function is positive and it's a concave up parabola, so there's no restrictions on x

for the range, look at the function f(x) = x^2+2x+3. that has range [2,inf) since at the vertex f(-1) = 2 and f goes to infinity as x goes to +/- infinity.
hence 1/f will have range (0,1/2] since at +/- infinity 1/f is gonna go to zero. taking square roots then just makes the range (0, 1/sqrt(2)]
thank you, I don't really understand your explanation for range could you clarify?

fastest way I've seen is
Domain: acknowledge function is a parabola -> all real numbers -> (-infin, infin)
Range: -b/2a for vertex -> find y value of vertex for max -> recognise as x approaches positive or negative infinity the fraction becomes closer to 0 (horizontal asymptote of 0) for min

@Shavi Masee honestly if u see something like this and it takes u more than 2 minutes i would just take my chances and graph it out for an answer. (might lose marks for working but worth it)
 

liamkk112

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thank you, I don't really understand your explanation for range could you clarify?

fastest way I've seen is
Domain: acknowledge function is a parabola -> all real numbers -> (-infin, infin)
Range: -b/2a for vertex -> find y value of vertex for max -> recognise as x approaches positive or negative infinity the fraction becomes closer to 0 (horizontal asymptote of 0) for min

@Shavi Masee honestly if u see something like this and it takes u more than 2 minutes i would just take my chances and graph it out for an answer. (might lose marks for working but worth it)
have u done the sketching 1/f(x) type of graphs stuff? it’s basically just using methods from there

what part do u want me to clarify about
 

NotBamboo

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have u done the sketching 1/f(x) type of graphs stuff? it’s basically just using methods from there

what part do u want me to clarify about
yeah i've done hyperbolas
this one is just a hyperbola with only positive y values
also this question was asked by shavi but i posted the thread for it

heres the bit im confused about, someone else mentioned a 2 in their working from another chat but i didnt understand it either

look at the function f(x) = x^2+2x+3. that has range [2,inf) since at the vertex f(-1) = 2 and f goes to infinity as x goes to +/- infinity.
hence 1/f will have range (0,1/2] since at +/- infinity 1/f is gonna go to zero.
taking square roots then just makes the range (0, 1/sqrt(2)]
 

liamkk112

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yeah i've done hyperbolas
this one is just a hyperbola with only positive y values
also this question was asked by shavi but i posted the thread for it

heres the bit im confused about, someone else mentioned a 2 in their working from another chat but i didnt understand it either

look at the function f(x) = x^2+2x+3. that has range [2,inf) since at the vertex f(-1) = 2 and f goes to infinity as x goes to +/- infinity.
hence 1/f will have range (0,1/2] since at +/- infinity 1/f is gonna go to zero.
taking square roots then just makes the range (0, 1/sqrt(2)]
okay it’s not quite a hyperbola, that would be something like 1/x . there’s a topic called “further graphs” or soemthing like that in ext 1 where u learn how to sketch 1/f(x) given any function f(x) that helps a lot with this question, but i’ll just explain what i talked abt

rather than looking at the function with the 1/sqrt of the parabola, let’s just take a look at the parabola itself and that will let us ascertain the information about 1/sqrt of the parabola
specifically let f(x)=x^2+2x+3 and g(x)=1/sqrt(f(x))
lets write down all of the information we know about f:
- for very large positive and negative x, the function goes to positive infinity, or the limit of f(x) as x goes to minus or plus infinity is infinity
- it has a turning point at the vertex (-1,2)
- the function is positive throughout its domain (which is all real x)
- the function has no zeroes

now consider the transformation h(x)=1/f(x) - using the information about f(x) we can talk about the properties of this function.
- firstly the domain is still all real x, as f has no zeroes, so h also has no vertical asymptotes
- h obviously has no zeroes
-since f goes to positive infinity as x goes to plus or minus infinity, h goes to 0 as x goes to plus or minus infinity (because 1/infinity = 0 in limits)
- the function is positive throughout its domain, since f is positive for all real numbers
-it has a turning point at (-1,1/2); this is inherited from the turning point of f(x), no need to differentiate h(x) (that’ll be explained in the module i was talking about)

since h goes to 0 at infinities, has a single turning point and doesn’t have any zeroes, we can conclude that the turning point now becomes a maximum rather than a minimum (this also happens because when you do 1/f(x), this changes all of the minimums into maximums). hence, the range must be (0,1/2], and the domain has been unchanged

now the final transformation is g(x)=sqrt(h(x)), remember that g(x) was what we were curious about
- when we take square roots, this doesn’t change behaviours at infinity
- h(x) is positive for all real numbers (has a positive maximum, no zeroes and goes to 0 at infinity) so g’s domain is also real numbers, as any number works under the square root
-g’s range is simply the square roots of h’s range, since it is defined on the same domain as h

hence the range of g is (0,1/sqrt(2)] and the domain is all real numbers

that was really long but hopefully that helps, it should make a lot more sense when you learn the topic i mentioned
 

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