Ive got as far as f'(x)= 2 - 18(x-4)^-2

the question is;

A function f is defined by f(x)= 2x + 3 + 18/(x-4), x is not equal to 4

## Winter Diet 2002 Paper 1 Question 10 can someone help please?

### loved-up-loon

Posted 15 May 2006 - 01:09 PM

Statinary points. (dy/dx=0)

nature table

And then see when its increasing.

nature table

And then see when its increasing.

### SncZ

Posted 15 May 2006 - 01:57 PM

the way i do it:

differentiate the function to get :

f'(x) = 2- 18/(x-4)

the clue to this question is realising that the function is increasing when the gradient is positive. ( which is why we differentiated the function )

2- 18/(x-4)

let 2- 18/(x-4)

2 - 18/(x-4)

2 = 18/(x-4)

cross multiply to get 2 (x-4)

expand the brackets

2(x

2x

2x

2(x

(x-7)(x-1) =o

x=7,x=1

You know that the function is 'u' shaped as the x

and they way i was taught, is that the only way to solve a quadratic inequation is to draw a graph... so draw a graph and mark on the points where the function crosses the x axis (0,1) (0,7) and the function is 'u' shaped so draw the graph passing through those two points, but label the axis x (horiz) and f'(x)(vert)

now... the function is +ve when the function is above the x axis which is when x<1 and when x>7 which is the same as 1<x<7

and finally remember to state:

as , y

- , y -

well thats the way i was told how to do them, i dunno how loved up loon does it .. why would you need the stationary points ?

differentiate the function to get :

f'(x) = 2- 18/(x-4)

^{2}which you gotthe clue to this question is realising that the function is increasing when the gradient is positive. ( which is why we differentiated the function )

2- 18/(x-4)

^{2}> 0let 2- 18/(x-4)

^{2}= o to find where it crosses the x axis2 - 18/(x-4)

^{ 2}= 02 = 18/(x-4)

^{ 2}cross multiply to get 2 (x-4)

^{2}= 18expand the brackets

2(x

^{2}-8x+16)=182x

^{2}-16x+32=182x

^{2}-16x+14=02(x

^{2}-8x+7)=0(x-7)(x-1) =o

x=7,x=1

You know that the function is 'u' shaped as the x

^{2}coefficient is +ve ( remember the old saying, if x^{2}is positive it has a smiley face i.e u shaped ? )and they way i was taught, is that the only way to solve a quadratic inequation is to draw a graph... so draw a graph and mark on the points where the function crosses the x axis (0,1) (0,7) and the function is 'u' shaped so draw the graph passing through those two points, but label the axis x (horiz) and f'(x)(vert)

now... the function is +ve when the function is above the x axis which is when x<1 and when x>7 which is the same as 1<x<7

and finally remember to state:

as , y

- , y -

well thats the way i was told how to do them, i dunno how loved up loon does it .. why would you need the stationary points ?

### loved-up-loon

Posted 15 May 2006 - 02:17 PM

Oh no sorry I was thinking of a different question. yeah that sounds about right, i think i did it the way i said before and just got totally confused! Ive just got a really bad memory! sorry!

### ad absurdum

Posted 15 May 2006 - 05:16 PM

SncZ's answer is perfect, but I thought I better point out that it is apparently not enough to say that the co-efficient of x

^{2}is positive, therefore it's a happy parabola and there is a minimum turning point, I lost a mark in my prelim for this. Apparently you have to take the derivative of the function (in this case the derivative of the derivative, i.e. the second derivative) and prove that it is a minimum turning point by use of nature table (or you could take yet another derivative and do it that way if you felt like giving the marker a sore head ). Maybe my teacher was just being harsh, but I guess it's always a good idea to put in more instead of less, if you've got time in the exam it's always best to make sure I think.### loved-up-loon

Posted 15 May 2006 - 05:29 PM

Say its (x-2) + 3

I no the x coordinate would be 2 and i think the y would be 3. But if your not sure, even though its only one mark, you could differentate it. but that could take up alot of time.

I no the x coordinate would be 2 and i think the y would be 3. But if your not sure, even though its only one mark, you could differentate it. but that could take up alot of time.

### ad absurdum

Posted 15 May 2006 - 05:30 PM

I was wondering, after expressing a quadratic in completed square form, how does one find the coordinates of the Turning Point? I know it's only worth one mark in the exam but I can't figure out what the formula is and it's bugging me.

For example if you have (x+a)

^{2}+ b, then your turning point will be the (-a,b) as -a makes the bracket equal to zero and +b is what is left after this.

### Joel

Posted 15 May 2006 - 05:37 PM

I'm not sure about a formula, but if you think of the the minimum turning point as the value which brings about the smallest value of y, then your value of x is going to be the value that makes the bracket zero (as any other value will make the bracket overall greater than zero). The value of y is then the constant added on.

For example if you have (x+a)

Yeah, i'm not having a problem with the x-coordinate. It's just the y-coordinate that i'm having trouble with.For example if you have (x+a)

^{2}+ b, then your turning point will be the (-a,b) as -a makes the bracket equal to zero and +b is what is left after this.Here's an example:

Find the coordinates of the TP of this function.

I thought this would be (-2, -11) but infact it is (-2, 11), can anyone explain this?

### ad absurdum

Posted 15 May 2006 - 05:45 PM

I'm not sure about a formula, but if you think of the the minimum turning point as the value which brings about the smallest value of y, then your value of x is going to be the value that makes the bracket zero (as any other value will make the bracket overall greater than zero). The value of y is then the constant added on.

For example if you have (x+a)

Yeah, i'm not having a problem with the x-coordinate. It's just the y-coordinate that i'm having trouble with.For example if you have (x+a)

^{2}+ b, then your turning point will be the (-a,b) as -a makes the bracket equal to zero and +b is what is left after this.Here's an example:

Find the coordinates of the TP of this function.

I thought this would be (-2, -11) but infact it is (-2, 11), can anyone explain this?

I'm sure that you are right, I think it may just be a mistake in the answers or something...Try putting x=-2 into the function, the ouput is -11 so I can't see where the +11 is coming from

### Joel

Posted 15 May 2006 - 05:50 PM

I'm sure that you are right, I think it may just be a mistake in the answers or something...Try putting x=-2 into the function, the ouput is -11 so I can't see where the +11 is coming from

Neither can I, the question was from a specimen question paper on the SQA website so I doubt there is a mistake. When I try other similar questions I keep on getting them wrong aswell.