How can I linearize that function?
f(x) = (x^3 - 2x + 4)^2 where a=1
(step by step)

- anonymous

How can I linearize that function?
f(x) = (x^3 - 2x + 4)^2 where a=1
(step by step)

- schrodinger

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- anonymous

do you know how to expand a function such as
\[\left( x ^{2} +3\right)^{2}\]
You multiply the first term by the first term, then 1st by 2nd, then 2nd by 1st, then 2nd by 2nd
Do you know what im talking about

- hba

@Skaematik Can't we use a formula there ?

- anonymous

What formula

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## More answers

- anonymous

L(x) = f(a) + f'(a)(x-a)

- hba

Formula to expand (x^2+3)^2
(a+b)^2=a^2+2ab+b^2

- hba

f(a)= (a^3 - 2a + 4)^2
then,
f'(a)= ?
Diffrentiate f(a) and find f'(a)

- anonymous

Why cant you just expand and take the natural log of both sides?

- hba

I am not sure about your input @Skaematik

- hba

@Pauloading Did you differentiate it ?

- anonymous

I imagine it's: f'(a) = 2.(3x^2 - 2.1 + 0), right?

- hba

I imagine that its f'(a) not f'(x)

- anonymous

This is an explanation I copied from a site
Starting with y = ax^b, we can take logs to any base c on both sides:
logc(y) = b logc(x) + logx (a)
So the graph of logc y vs. logcx is linear, with slope b and y-intercept equal to logc a.

- hba

@Pauloading do you follow @Skaematik Method ?

- anonymous

Not yet :x

- hba

@Pauloading Do you want to follow my method ?

- anonymous

I'ts sound like greek to me :'(
I'm really bad on it.

- g152xx

if it is f-1(a) when a = 2...
set the first equation equal to 2, which will happen when x = 1.
So, f-1(2) = 1.
Now if you want f-1'(x), you have:
= 1 / f'(f-1(x))
so 1/ f'(1)
find the derivative of f(X):
3x^2 + 2
so,
answer = 1/(3(1) +2) = 1/5.
Now just replace a = 1 instead of a = 2. Easy.

- hba

@g152xx I hope you can help @Pauloading ,i gotta go workout ?

- anonymous

Imagine that I've a new function:
f(x) = x^3 a=1
and also I try to apply that formula --> L(x) = f(a) + f'(a)(x-a)
L(x) = f(a^3) + f'(3a^2)(a^3-1)
L(x) = (1^3) + f'(3.1^2)(1^3-1)
L(x) = 1 + 3.(0)
L(x) = 1
Is it right?

- g152xx

@hba I'll surely try!

- anonymous

Thanks @hba :)

- hba

Do you know that a linear function has a point (x1,y1) (Let) and a slope m (Let),Anyways Thanks a lot @g152xx And @Pauloading Your'e welcome.

- g152xx

@Pauloading
It sounds right 2me. Unless you wanted the answer in the form L(x) = f(a)+f'(a)(x-a) instead of simplifying it.....

- g152xx

@Pauloading Bump the question! Let's help him ppl! C'mon!

- hartnn

we have the formula, L(x) = f(a) + f'(a)(x-a) why don't we use it directly ??
a=1
f(1) = 9
f'(1)=6
L(x) = 9+6(x-1)
= 6x +3

- anonymous

How did u get 6 at f'?

- hartnn

could you find f'(x) =... ?

- hba

I asked her the same thing before :D

- anonymous

I got 2.(3x^2 - 2)

- hartnn

no, thats not correct... u need chain rule.

- anonymous

hm....got it

- hartnn

ok, what u got f'(x) =... ?

- hba

The answer would be 2 (-2 + 3 a^2) (4 - 2 a + a^3)

- hartnn

let me verify it for you....

- hartnn

Imagine that I've a new function:
f(x) = x^3 a=1
and also I try to apply that formula --> L(x) = f(a) + f'(a)(x-a)
L(x) = f(a^3) + f'(3a^2)(a^3-1) <-----INCORRECT
L(x) = (1^3) + f'(3.1^2)(1^3-1)
L(x) = 1 + 3.(0)
L(x) = 1
L(x) = (1^3) + (3.1^2)(x-1)
=3+3(x-1)
=3x

- anonymous

Now I understood it '-'
Thanks everyone :D

- hba

d/da((a^3-2 a+4)^2)
Using the chain rule, d/da((a^3-2 a+4)^2) = 2 u ( du)/( da), where u = a^3-2 a+4 and ( du^2)/( du) = 2 u:
2 (a^3-2 a+4) (d/da(a^3-2 a+4))
Differentiate the sum term by term and factor out constants:
2 (a^3-2 a+4) (d/da(a^3)-2 (d/da(a))+d/da(4))
The derivative of 4 is zero:
2 (a^3-2 a+4) (d/da(a^3)-2 (d/da(a))+0)
The derivative of a is 1:
2 (a^3-2 a+4) (d/da(a^3)-2 1)
The derivative of a^3 is 3 a^2:
Answer: |
| 2 (a^3-2 a+4) (3 a^2-2)

- hartnn

was that ^ copied ?

- hartnn

from wolf ?

- hba

Yeah :D

- hartnn

gotcha.....

- hba

I had to show her how to do it :)

- hba

him*

- hartnn

okay.
welcome paul ^_^

- hba

Paulo*

- anonymous

Thanks >.<

- hba

Your'e Welcome.

- hartnn

ok,\[ \begin{array}l\color{red}{\text{P}}\color{orange}{\text{a}}\color{#e6e600}{\text{u}}\color{green}{\text{l}}\color{blue}{\text{o}}\color{purple}{\text{ }}\color{purple}{\text{C}}\color{red}{\text{e}}\color{orange}{\text{c}}\color{#e6e600}{\text{í}}\color{green}{\text{l}}\color{blue}{\text{i}}\color{purple}{\text{o}}\color{purple}{\text{}}\end{array} \]

- anonymous

LoL :D

- hba

\[\Huge{\bf{\color{blue}{O}\color{red}{K}}}\]

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