#A level maths help

61 messages · Page 1 of 1 (latest)

zenith ridge
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How would i find the range of part b . ive done part a and got 3(x-2)^2+8 but idk where to go from there.

buoyant rune
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obviously the quadratic in a is the denominator in b so g(x) you can write as

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1/(3(x-2)^2 + 8)

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right

zenith ridge
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yep

buoyant rune
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now that (x-2)^2

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what can you tell me about the range of that

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what can you tell me about the output of it

zenith ridge
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that it has to be positive

buoyant rune
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yep or 0

zenith ridge
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yeah

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if x=2

buoyant rune
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now 3 times that obviously wont make a difference

zenith ridge
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yeah

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oh so the lowest is 1/8

buoyant rune
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exactly

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and it can be 1/(8 + positive)

zenith ridge
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so its g(x) > or euqla to 1/8?

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equal

buoyant rune
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ah not quite so

zenith ridge
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ok

buoyant rune
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lowest actually isnt right i was wrong

zenith ridge
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thats fine

buoyant rune
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so if you made the 3(x-2)^2 positive

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say instead it was like 3

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your g(x) would be 1/11 right

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so smaller than your 1/8

zenith ridge
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say instead what was 3?

buoyant rune
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your 3(x-2)^2

zenith ridge
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oh ok yhyh

buoyant rune
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just pretend you put in a value so it was 3

zenith ridge
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ok

buoyant rune
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then your results gonna be 1/11

zenith ridge
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yeah

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sa

buoyant rune
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smaller than your 1/8

zenith ridge
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yeap

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yep

buoyant rune
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what if you plugged in a larger number

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like as you make 3(x-2)^2 bigger and bigger

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whats gonna happen to your 1/(3(x-2)^2 + 8)

zenith ridge
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that itll just get smaller and smaller

buoyant rune
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yep and importantly

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itll 'approach' a number

zenith ridge
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0?

buoyant rune
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like it wont go off to - infinity

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perfect

zenith ridge
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ah ok

buoyant rune
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itll approach 0

zenith ridge
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so is the lim thing?

buoyant rune
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so g(x) is at most 1/8

zenith ridge
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ok

buoyant rune
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but also greater than 0

zenith ridge
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ah

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ok i see

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so its greater than zero but less than or equal to 1/8?

buoyant rune
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perfect

zenith ridge
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ok cool that makes sense now

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thanks a lot

buoyant rune
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no worries!