Math Geniuses..come in here plz..(derivitives)

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shudnt_have

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If I am asked to find the constants a, b, c, d such that the graph of
f(x)= ax^3 + bx^2 + cx+ d has horizontal tangent lines at the points (-2, 1) and (0, -3).

I know though, the first thing I would do is, find the derivative of the function..."f(x)= ax^3 + bx^2 + cx+ d "

which would be...

f' (x) = 3 ax^2+ 2 bx+ c
then sub the value of x? x = (-2) into the last equation..
which will equal to
12 a- 4b+ c

I got no clue onwards..:D


and I have another q.


Given h= f 0 g, g(3)=7, g'(3)=4, f(2)=4, f'(7)=-6.
now how do I determine the h' (3)???

again i am half way through the answer...

I think thinking of solving it with product rule??
h(x)= f(g) x)) h(x)= f' (g(x) g'(x)

h' (x)= f' (g (3)= g'3= f' (7) (4) = (-6) (-4) = -24
 
Sis I don't have a clue!! . It's been a long time since my A-level math's. Perhaps you can give me a site to brush up my math's on that level.
 
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Salam,

I haven't done this yet, but if you provide the link (like br skillganon mentioned), then, I will have a go. It'll be fun!

Wasalam
 
Again been a long time since I did this kind of thing...

I got

a: 1
b: 3
c: 0
d: -3

Disclaimer: I might be wrong :p ... I'll have another look
 
:sl:

Remember the gradient of a tangent line is the same as the gradient of the curve at the same point. Now, since both the tangents are horizontal lines (from the question), the gradients of both lines are then taken to have the value 1. This means then that the gradients of the curve at points (-2,1) and (0,-3) are also both 1:

Gradient of curve at [-2,1]: 12a - 4b + c = 1
Gradient of curve at [0,-3]: c = 1
Click to expand...

I don't think the gradient of a horizontal line is 1. A gradient of 1 is like the y=x graph. I think gradient of a horizontal line is 0.

If you recalculate with this change you should get same anwser as me.

Now for the second question, no idea, I don't even understand the notation.

what is f 0 g ??
 
Just had a thought, those two coordinates given are stationary points on the curve and at stationary points dy/dx is always zero because the sign of the curve has to change. eg from minus to plus or plus to minus, therefore passing through zero.

Eg.
fig.gif
 
Salam,

Br Alpha Dude and MH-UK could you kindly recommend a website that includes this topic and other AS topics. Please? This sounds really fun (challenging). I tried to understand the exercises on the links that I found, but they weren't very good.

Wasalam
 
salams

bro alpha has the correct values of b,c and d, but a=+1

here's my working

f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
f'(-2) = 0 and f'(0) = 0

eq1. f'(-2) = 0 = 12a - 4b + c
eq2. f'(0) = 0 = c
eq3. f(-2) = 1 = -8a + 4b + d
eq4. f(0) = -3 = d

thrfore, c=0, d=-3 from eq2 and eq4 respectively.

sooo.. we can deal with these equations now (i subbed in c=0 and d=-3 into eq1 and eq3.

eq1. f'(-2) = 0 = 12a - 4b
eq3. f(-2) = 1 = -8a + 4b - 3

i simplified those two equatons to get

eq1. f'(-2) = 0 = 3a - b
eq3. f(-2) = 0 = -2a + b - 1

now, we have simple case of simultaneous equations. from eq1. we get
eq1 a = b/3

we can sub a=b/3 into eq3 and get

eq3 0 = -2b/3 + b -1

from that we get,

b = 3

therefore

now we can substitute b=3 into eq1 and get value for a

eq1 3a-3 = 0
a= 1

hence a=1, b=3, c=0, d=-3

now ill take a look at the 2nd problem n get bak2u later insahlah.. salams

ps: i'm not a math genuis, i suck at it.. i learnt calculus from rajuman's thread about 3=2.
 
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salams sis

as for the second problem.. we solve it using the chain rule

basically this is how the chain rule goes.

for the function h = f(g(x))

h'(x) = f'(g(x)) * g'(x)

sooo... therefore we can say

h'(3) = f'(g(3)) * g'(3) = f'(7) * 4 = -6 * 4 = -24

so yep ur right.. but that's called chain rule not product rule.
salams
 
no prob :D














lol messing.. i proved you right too, you where correct in 3 of the 4 cases for a,b c, d :)
 
Hola Math Geniuses!

Now I need your help... this is very difficult. If I have 10,000 add 1% onto that then put the 1% back into the 10,000 and then repeat the process with 10,100 and continue like that for about 51 more times what would my end result be?
 
heya jayda... could you explain it perhaps by presenting an example

i didnt quite understand the pattern.. but i got a feeling its to do with sequences and series..

take carea ll the best!
 
Jayda said:
Hola Math Geniuses!

Now I need your help... this is very difficult. If I have 10,000 add 1% onto that then put the 1% back into the 10,000 and then repeat the process with 10,100 and continue like that for about 51 more times what would my end result be?
Click to expand...

are you adding 1% onto 10 000 to get 10 100
then also putting 1% back into the 10 000 which also sort of is 10 100
but how can we perform that operation twice
except that it is implicit in repeating the process with 10 100 that adding and putting back are the same
so 52 operations of add/put back 1% of 10 000

= a pattern that is definitely beyond the number at which I would assume that it continues in the same fashion as the earliest pattern which shows

eg. I might have said, if the question was only 41 more operations that the answer is like 14242

BUT after 42 the pattern most likely goes askew. But my self ain't going to sit down with pen and paper to make those calculations right now to know for certain. Fourty two has a stop probability higher than other numbers, and that ain't no superstition but a verifiable fact of logic.

Look at the safe curve by: 9>13>18>26>36>42.

However as far as derivatives and differentials go you lot are all far surpassing me. I learned late as a teenager by accident of a change of teacher and then over relied upon my father who is as often as not less accurate than I am (shhh). Then I had to relearn at University in which I remember distinctly looking at the measuring of a tangent with utter horror: "they can't do that!" So we stop at the fourty two, because the thirty six was bad enough, and the top of the range is the twenty six in this sum. (I hope the Mahdi sorts it all out.)

True it is often difficult for believers to comprehend calculus becasue in the real world what many of the set examples which are used are able to be deciphered into is that need for Jahannam to exist! There is only one example of calculus which is representative of Qur'an, and it is which ever example it is which causes the mind to orient to Revelations Isa made to St John the Evangelist.

But that is the utter limit of mathematical genius, and what is the meaning of the word genius anyhow? Is it derived from "genesis"? or at least having the same greek or latin root?
 
MH-UK said:
:sl:



I don't think the gradient of a horizontal line is 1. A gradient of 1 is like the y=x graph. I think gradient of a horizontal line is 0.

If you recalculate with this change you should get same anwser as me.

Now for the second question, no idea, I don't even understand the notation.

what is f 0 g ??
Click to expand...

When is the horizon a horizontal line?;D
 
But I don't get why there was a series of coincidences in respect of these posts aligned with the numbers:
111
84
5

Which are an interesting sequence.
 
^ lol sis i think he's referring to f(g) when he says fog.... basically suppose you have a function

g(x) = x^2

and another function f,

f(x) = 3x+2

f(g) (fog) basically means you take g(x) and substitute x for g(x) in f(x)... so f(g) becomes:

f(g) = 3(x^2) + 2

I don't think the gradient of a horizontal line is 1. A gradient of 1 is like the y=x graph. I think gradient of a horizontal line is 0.
Click to expand...

ya ur right... another way to look at it is.. if you take the hoirzontal line

y = 3

and take the derivative of that .... it would be zero. (derivative of a constant is zero).

or.. analysing it fromt he defintion of gradietn = 'rise/run'.... the rise of a horizontal line is zero.. so zero divided by one is zero

take care all the best
 
:sl:

lolwatever, you're right, and from that argument you can say that a vertical line has a "run" of 0 and therefore you would be dividing by 0, which gives an infinite gradient I imagine....
 

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