M.Johnson2111
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Hi,

If g (x) > 28 find all values for x when g (x) = 4x - 7

I've found x > 35/4

The question is worth four marks so feel as though I need to do more or have misread the question?
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purplekitten42
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What you did seems right but I agree that it seems easy for four marks. Is that the exact wording of the question?
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M.Johnson2111
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M.Johnson2111
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(Original post by purplekitten42)
What you did seems right but I agree that it seems easy for four marks. Is that the exact wording of the question?
Hi thanks for the reply. I've uploaded a copy of the full question. I'm also struggling with part d?
Do i need to find the inverse function of h (x) , make this equal -1/2 to solve
Or let -1/2 be x and sub into h (x) to solve?
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purplekitten42
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(Original post by M.Johnson2111)
Hi thanks for the reply. I've uploaded a copy of the full question. I'm also struggling with part d?
Do i need to find the inverse function of h (x) , make this equal -1/2 to solve
Or let -1/2 be x and sub into h (x) to solve?
In part b there will also be values of x that satisfy the equation for the g (x) = (x - 2)^2 + 1 which is why the question is four marks.

For part d, you have the right idea to find the inverse function of h (x) and make it equal to -1/2.
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RDKGames
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(Original post by M.Johnson2111)
Hi thanks for the reply. I've uploaded a copy of the full question. I'm also struggling with part d?
Do i need to find the inverse function of h (x) , make this equal -1/2 to solve
Or let -1/2 be x and sub into h (x) to solve?
If you sketch g(x), you would notice that the quadratic branch takes the function beyond $28$ just like the right linear branch. Hence, you need to also solve

(x-2)^2 +1 > 28 for x\leq 2.

For part d, sure you can find inverse h^{-1} and then find x ... but a nice little cheat is to realise that if

h^{-1}(x) = -1/2,

then

x = h(-1/2)

since you can just apply h to both sides.
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M.Johnson2111
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(Original post by RDKGames)
If you sketch g(x), you would notice that the quadratic branch takes the function beyond $28$ just like the right linear branch. Hence, you need to also solve

(x-2)^2 +1 > 28 for x\leq 2.

For part d, sure you can find inverse h^{-1} and then find x ... but a nice little cheat is to realise that if

h^{-1}(x) = -1/2,

then

x = h(-1/2)

since you can just apply h to both sides.
Thank you
For part d I found x = 29/4 but since the first part is squared could another x value be x = -29/4?
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RDKGames
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(Original post by M.Johnson2111)
Thank you
For part d I found x = 29/4 but since the first part is squared could another x value be x = -29/4?
No.

Inverses exist when they are one-to-one, and if what you are saying would be true, it would mean that the inverse function is a many-to-one mapping... this is not possible.

Also, in the first part it is indeed squared, but the domain of this square is restricted to x \leq 2 which makes it a one-to-one function. No way are you going to get two different values!


And one more thing ... the domain of the inverse function is the range of the original function ... but the range of the original function is \geq 1, so how can you have -29/4 as an input into the inverse function?
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M.Johnson2111
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(Original post by RDKGames)
No.

Inverses exist when they are one-to-one, and if what you are saying would be true, it would mean that the inverse function is a many-to-one mapping... this is not possible.

Also, in the first part it is indeed squared, but the domain of this square is restricted to x \leq 2 which makes it a one-to-one function. No way are you going to get two different values!


And one more thing ... the domain of the inverse function is the range of the original function ... but the range of the original function is \geq 1, so how can you have -29/4 as an input into the inverse function?
Thank you!
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