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y = f(x) and its inverse

OCR Past Paper June 2005 - Question 9 - part iii :

curve f(x) = root (mx + 7) -4

It is given that f(x) and its inverse do not meet - determine the possible value of m.

The answers from the markscheme is 2 < m < 14, but I cannot get this?

Any help appreciated ...
Reply 1
Avatar for sciortino
sciortino
Letting the inverse, ((x+4)^2-7)/m, equal f(x) produces x=(sqrt(m^2-16m+28)+m-8))/2. This has no solutions when (m^2-16m+28)<0, viz. (m-14)(m+2)<0. Hence 2<m<14.
Reply 2
Avatar for sciortino
sciortino
Sorry, meant (m-2).
Reply 3
Avatar for Charries
Charries
OP
3
I still need help as I cannot make the jump you have ...

I could look at the inverse as you suggest (but actually, I thought I could just set the original function to x)

... but doing the inverse ...

I have inverse as y = ((x+4)^2-7)/m

Let y = x, then x = ((x+4)^2-7)/m

so mx = (x+4)^2 - 7
so mx = x^2 + 8x + 16 -7
so x^2 + (8-m)x + 11

so discrim is (8-m)^2 - 44 <0
so m^2 - 16m + 64 - 44 < 0
so m^2 - 16m - 20 < 0

which is not m^2 -16m +28 ???????? - i am lost ...
Reply 4
Avatar for V.S
V.S
correction "mx = x^2 + 8x + 16 -7" equals m^2 +(8-m)x +9

using b^2 - 4ac discriminant
(8-m)^2 -36
64 - 16m + m^2 -36
which equals m^2-16m+28
Reply 5
Avatar for sciortino
sciortino
Okay, following your idea....

f(x)=x yields mx+7=(x+4)^2=0

mx+7=x^2+8x+16=0

x^2+(8-m)x+9=0

This must have no solutions, so (8-m)^2-36<0

m^2-16m+28<0 and the result follows as before.
Reply 6
Avatar for Charries
Charries
OP
3
my god i am a fool - i tired it about 8 times and still made the same error - thanks ...

... walks away quite embarrassed ...
Reply 7
Avatar for sciortino
sciortino
Ignore my extra =0s.
Reply 8
Avatar for Charries
Charries
OP
3
... still got a bit of a problem - we say that :

m^2-16m+28<0 and the result follows as before.

so (m-2)(m-14) < 0

doesn't this get us, m < 2 , m < 14 ?

and not, 2 < m < 14?
Reply 9
Avatar for james.h
james.h
11
Charries
... still got a bit of a problem - we say that :

m^2-16m+28<0 and the result follows as before.

so (m-2)(m-14) < 0

doesn't this get us, m < 2 , m < 14 ?

and not, 2 < m < 14?

There are different ways of thinking about this, but I find the following the most intuitive:

m2−16m+28\displaystyle m^2 - 16m + 28 is a positive parabola, with a minimum turning point.

You've found its roots to be m=2 and m=14.
Thus, the minimum turning point must lie between these points, and below the x-axis.
And hence there's your answer: 2 < m < 14. :smile:

(Hope that makes sense. :o:)

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