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Shavi Masee
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Ty bestiee I probably should have started a new thread when I posted mb T-T
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)
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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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
1/f(x)
i forgor what it's called feel free to correct me lol
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why tf you doing the stupid ass test method? Just sketch a parabola using the zeros after factorising
good luck factorising (x^2 + 2x + 3)
funny answers
Shavi Masee
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I tried completing the square, that didn't help at all T-T
liamkk112
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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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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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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