double differentiation question (1 Viewer)

[micuzzo]

can anyone please help on this question... i no its easy but i havent as of yet experienced one as such

f''(x)=8 and f'(x)=0 and y=3 when x=1 find equation of y in terms of x


Thanks you, i

 

[gurmies]

tell me, is it:

y = 4x^2 - 8x + 7

I've also never done one of these before, so it's quite possible that i'm incorrect in my approach.

 

[micuzzo]

yes... but how to do it

 

[gurmies]

Oh, beautiful. Well, I though of it like this:

Not too sure why, but I assumed it was a quadratic..(hmm..this is really dodgy), so I called it:

y = ax^2 + bx + c

y' = 2ax + b

y'' = 2a

Now, you said, y'' = 8. Therfore:

2a = 8

a = 4.

Now, back to the y'

y' = 8x + b [I subbed in a= 4]

When x = 1, y' = 0

b + 8 = 0

b = -8

Now, back to the initial equation,

y = 4x^2 - 8x + c

Since, when x = 1, y = 3

3 = 4 - 8 + c

c = 7

Therefore, the equation is 4x^2 - 8x + 7. But that's crap, I wonder why exactly it's a quadratic.. for some reason I thought there was only 1 turning point..

 

[micuzzo]

Thanks... sorry bout pm... u must have answered it straight after i refreshed... seems right,,, thank neway

 

[gurmies]

oh no, that's ok. You can pm me any time Not only does it help you, but it can also test my own knowledge, which is always desired. I'm still not satisfied however, in assuming that there was ONLY 1 turning point, in other words, a quadratic. Maybe somebody could dispell any doubts of mine..?

 

[micuzzo]

hey what 4U book do you use?

 

[gurmies]

I use a variety.

For school, I'm currently using a 4 unit Coroneos book for complex numbers, which apparently has very good complex number questions. After that, we'll be using fitzpatrick for the rest of the topics.

With my tutor, i'm using the cambridge book (arnold). However, he has also given me a university book from hong kong, which is better than any book here in Australia. It has some of the hardest and most interesting questions.

 

[micuzzo]

woh... good luck with that... i currently use Terry Lee... which i think is a bit over the top... Fitz seems ok... but judging from complex numbers it is easier than terry lee

 

[gurmies]

Well, to be honest, a lot of my friends who have just finished their exams, claim that terry lee is very good. bored_of_sc made a post in the mathematics extension 2 section, about a question from the terry lee book which I found very very interesting. If it has many of those questions, then it's definitely a good book. Advanced material won't hurt

 

[tommykins]

can anyone please help on this question... i no its easy but i havent as of yet experienced one as such

f''(x)=8 and f'(x)=0 and y=3 when x=1 find equation of y in terms of x


Thanks you, i

f'(x) = int. 8 = 8x + c, when x = 1, f'(x) = 0.
8 + c = 0
c = -8

f'(x) = 8x-8
f(x) = int . f'(x) = 4x^2 - 8x + c x = 1, y = 3

3 = 4 - 8 + c
c-4 = 3
c = 7.

Thus y = 4x^2 - 8x + 7

 

[3.14159potato26]

f''(x)=8 and f'(x)=0 and y=3 when x=1 find equation of y in terms of x

Well, I dunno about the previous solution (not that I have anything against it, really) but here's how I look at it:
To determine the coefficients of terms in an nth degree polynomial, you need at least (n+1) distinct equations in terms of the coefficients.
This rule is quite obvious. For example, if the polynomial was of degree zero, you need one y-value, e.g. the polynomial is y = c, so you need the y-value to get the value of c.
For the question to be solvable, and since the 2nd differential is non-zero, therefore the equation is a 2nd degree polynomial, i.e. is a quadratic polynomial because it gives 3 equations in terms of the coefficients.
Therefore, taking f(x) = ax^2 + bx + c, it gives the equations:
a + b + c = 3
2a + b = 0
2a = 8 or a = 4.
which can be rewritten as (rearranging the order of the terms):
{ 1 1 1 3 }
{ 0 1 2 0 }
{ 0 0 1 4 }
which is a 3*4 matrix, which can be solved using backsubstitution to give:
{ 1 0 0 7 }
{ 0 1 0 -8 }
{ 0 0 1 4 }
that is, a = 4, b = -8 and c = 7.
Therefore, y = 4x^2 - 8x + 7.

 

[Timothy.Siu]

ur supposed to do tommykins way

 

[shaon0]

Well, I dunno about the previous solution (not that I have anything against it, really) but here's how I look at it:
To determine the coefficients of terms in an nth degree polynomial, you need at least (n+1) distinct equations in terms of the coefficients.
This rule is quite obvious. For example, if the polynomial was of degree zero, you need one y-value, e.g. the polynomial is y = c, so you need the y-value to get the value of c.
For the question to be solvable, and since the 2nd differential is non-zero, therefore the equation is a 2nd degree polynomial, i.e. is a quadratic polynomial because it gives 3 equations in terms of the coefficients.
Therefore, taking f(x) = ax^2 + bx + c, it gives the equations:
a + b + c = 3
2a + b = 0
2a = 8 or a = 4.
which can be rewritten as (rearranging the order of the terms):
{ 1 1 1 3 }
{ 0 1 2 0 }
{ 0 0 1 4 }
which is a 3*4 matrix, which can be solved using backsubstitution to give:
{ 1 0 0 7 }
{ 0 1 0 -8 }
{ 0 0 1 4 }
that is, a = 4, b = -8 and c = 7.
Therefore, y = 4x^2 - 8x + 7.

wow you used matrices. That's pretty good.

 

[lyounamu]

Well, I dunno about the previous solution (not that I have anything against it, really) but here's how I look at it:
To determine the coefficients of terms in an nth degree polynomial, you need at least (n+1) distinct equations in terms of the coefficients.
This rule is quite obvious. For example, if the polynomial was of degree zero, you need one y-value, e.g. the polynomial is y = c, so you need the y-value to get the value of c.
For the question to be solvable, and since the 2nd differential is non-zero, therefore the equation is a 2nd degree polynomial, i.e. is a quadratic polynomial because it gives 3 equations in terms of the coefficients.
Therefore, taking f(x) = ax^2 + bx + c, it gives the equations:
a + b + c = 3
2a + b = 0
2a = 8 or a = 4.
which can be rewritten as (rearranging the order of the terms):
{ 1 1 1 3 }
{ 0 1 2 0 }
{ 0 0 1 4 }
which is a 3*4 matrix, which can be solved using backsubstitution to give:
{ 1 0 0 7 }
{ 0 1 0 -8 }
{ 0 0 1 4 }
that is, a = 4, b = -8 and c = 7.
Therefore, y = 4x^2 - 8x + 7.

Wow, that's a great way to look at it. But I prefer the tommykin's

 

[Continuum]

Well, I dunno about the previous solution (not that I have anything against it, really) but here's how I look at it:
To determine the coefficients of terms in an nth degree polynomial, you need at least (n+1) distinct equations in terms of the coefficients.
This rule is quite obvious. For example, if the polynomial was of degree zero, you need one y-value, e.g. the polynomial is y = c, so you need the y-value to get the value of c.
For the question to be solvable, and since the 2nd differential is non-zero, therefore the equation is a 2nd degree polynomial, i.e. is a quadratic polynomial because it gives 3 equations in terms of the coefficients.
Therefore, taking f(x) = ax^2 + bx + c, it gives the equations:
a + b + c = 3
2a + b = 0
2a = 8 or a = 4.
which can be rewritten as (rearranging the order of the terms):
{ 1 1 1 3 }
{ 0 1 2 0 }
{ 0 0 1 4 }
which is a 3*4 matrix, which can be solved using backsubstitution to give:
{ 1 0 0 7 }
{ 0 1 0 -8 }
{ 0 0 1 4 }
that is, a = 4, b = -8 and c = 7.
Therefore, y = 4x^2 - 8x + 7.

I prefer tommykins method, haha. Although matrices look cooler.