Higher Maths Question

Call Me Jimmy

hey i was going over differentiation and while doing one of the questions from the mock I got stuck on the last part of a (c) part and i'm wondering if any1 can clear this up. Basically it's asking me to find the equation of the tangent to the graph of a function that i just proved had no turning points and not points of inflection. So if the graph has no turning points and no points of inflection, does that not mean its a straight line? like slanty but straight? and if so then i have to find a point on it and a slope? or something like that? I am not sure if i have made much sense but if any1 can help i'de appreciate it

*Angel*

Call Me Jimmy wrote:

like slanty but straight?

Diagonal maybe?

Yes well you could put the full question up cos it's kinda hard to tell, but you already got the first derivative (the slope), so you have a point on the line that you could sub in to get the slope.

Perhaps you were given a range and it didn't have turning points or a point of inflection in that range?

I'd like to see the question too.

Call Me Jimmy

k i wasnt posting it originally coz i was concerned mainly with the question of the line with no asymptotes and no inflection being a (diagonal ) line, ok well heres the question

f(x) = x
____
x-3 where x E R and x doesnt = 3

(i) Find the equations of the asymptotes of the graph f (check)
(ii)Prove that the graph of f(x) has no turning points and no points of inflection (check)
(iii) Given that 3x+y=k is a tangent to the graph of f where k is a real number, find two possible values of k.

*Angel*

yeah so from 3x + y = k you can figure out that the slope is -3

f'(x) = -3/(x-3)^2

- 3 = -3/(x-3)^2

(x - 3)^2 = 1
x^2 - 6x + 8 = 0
(x - 4)(x - 2) = 0
x = 4, 2

y = x/(x - 3)

x = 4, y = 4
x = 2, y = -2

subbing into 3x + y = k

k = 16, 4

oq4v3ht0u76kf2

Roysh then, let's get cracking...

f(x) = x / (x -3)
f'(x) = -3 / (x - 3)^2

3x + y = k
y = -3x + k

Therefore the slope, m = -3    [due to y = mx + c]

f'(x) = m
- 3 / (x - 3)^2 = -3
Simplifies to
x^2 - 6x + 8 = 0
(x - 4)(x - 2) = 0
x = 4 OR x = 2

x = 4
f(4) =  y = 4

x = 2
f(2) = y = -2


Two values for k: 
3x + y = k

x = 4, y = 4
3(4) + 4 = 16 #

x = 2, y = -2
3(2) - 2 = 4 #

So k = 16 or k = 4

Call Me Jimmy

thanks but in the first line, after you establish that the slope of tangent is -3 then you say
f'(x) = -3/(x-2)^2 but THEN
you put the tangent slope = to the graph f slope no?
-3 = -3/(x-2)^2 ? how is this so? if the (diagonal) line has a tangent, then surely the tangent slope cant equal the line slope? i must be missing something sorry

the smiley one

*Angel* wrote:

yeah so from 3x + y = k you can figure out that the slope is -3

f'(x) = -3/(x-3)^2

- 3 = -3/(x-3)^2

(x - 3)^2 = 1
x^2 - 6x + 8 = 0
(x - 4)(x - 2) = 0
x = 4, 2

y = x/(x - 3)[/B]
x = 4, y = 4
x = 2, y = -2

subbing into 3x + y = k

k = 16, 4

Can you use that fact, because surely that is the x value of a point on the curve, not neccessarily on the line? Therefore subbing in 2 and 4 will get you the corresponding points on the curve......not the line?!
I get really confused about stuff like this......



Anyone like to explain???!!!!

the smiley one

Call Me Jimmy wrote:

thanks but in the first line, after you establish that the slope of tangent is -3 then you say
f'(x) = -3/(x-2)^2 but THEN
you put the tangent slope = to the graph f slope no?
-3 = -3/(x-2)^2 ? how is this so? if the (diagonal) line has a tangent, then surely the tangent slope cant equal the line slope? i must be missing something sorry

f'(x) is always the slope of the tangent to the curve - that is it's definition
Not sure if you are asking that though...but anyway

oq4v3ht0u76kf2

Are you sure it's a diagonal line? See my attachment. That curve has no turning points and no points of inflection.

It clearly isn't a diagonal line. A fraction like that would be as is in the diagram above.

*Angel*

Call Me Jimmy wrote:

thanks but in the first line, after you establish that the slope of tangent is -3 then you say
f'(x) = -3/(x-2)^2 but THEN
you put the tangent slope = to the graph f slope no?
-3 = -3/(x-2)^2 ? how is this so? if the (diagonal) line has a tangent, then surely the tangent slope cant equal the line slope? i must be missing something sorry

you're given 3x + y = K, (y = mx + c), m = -3

f'(x) of the 'curve' is the slope of a tangent to a curve

therefore they are equal

Call Me Jimmy

oh yeah bob, ur absolutely right. Lol wasnt even thinking straight, now im good, thanks for the help people, much appreciated!

oq4v3ht0u76kf2

No worries! Best of luck in the exam man!

f'(x) of the 'curve' is the slope of a tangent to a curve

Indeed. The whole basis of that question was to see if you know that.

*Angel*

the smiley one wrote:

f'(x) is always the slope of the tangent to the curve - that is it's definition
Not sure if you are asking that though...but anyway

you've found out that x is a point on the curve where a line is tangental, therefore the corresponding y value will be on the curve

Call Me Jimmy

yeah angel and aristotle i realise i just wasnt thinking right, thanks

the smiley one

*Angel* wrote:

you've found out that x is a point on the curve where a line is tangental, therefore the corresponding y value will be on the curve

So does that mean that the two points you have found are on both the line and the curve?

If you see the [URL=https://us.v-cdn.net/6034073/uploads/attachments/8386/14021.PNG[/URL] , you will see that the curve splits into opposite sides of the X and Y axis (I can't remember what that's called at the mo ). So you could have a line on either one of those.

oq4v3ht0u76kf2

It's a rational function... that is f(x) = g(x) / h(x)

The only two on our course are f(x) = a / (x + b) and f(x) = x / (x + b)

The horizontal/vertical lines that the curve never meets are called the asymptotes.

Vertical asymptote: the denominator = 0 (i.e. x + b = 0, x = -b)
Horizontal asymptote: y = lim[x -> infinity] f(x)

rational

That's the word I couldn't think of!

oq4v3ht0u76kf2

You don't do your nickname justice man! G'luck!

LoL! Cheers.