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

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SkaematikBest ResponseYou've already chosen the best response.0
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
 one year ago

hbaBest ResponseYou've already chosen the best response.0
@Skaematik Can't we use a formula there ?
 one year ago

PauloadingBest ResponseYou've already chosen the best response.0
L(x) = f(a) + f'(a)(xa)
 one year ago

hbaBest ResponseYou've already chosen the best response.0
Formula to expand (x^2+3)^2 (a+b)^2=a^2+2ab+b^2
 one year ago

hbaBest ResponseYou've already chosen the best response.0
f(a)= (a^3  2a + 4)^2 then, f'(a)= ? Diffrentiate f(a) and find f'(a)
 one year ago

SkaematikBest ResponseYou've already chosen the best response.0
Why cant you just expand and take the natural log of both sides?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
I am not sure about your input @Skaematik
 one year ago

hbaBest ResponseYou've already chosen the best response.0
@Pauloading Did you differentiate it ?
 one year ago

PauloadingBest ResponseYou've already chosen the best response.0
I imagine it's: f'(a) = 2.(3x^2  2.1 + 0), right?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
I imagine that its f'(a) not f'(x)
 one year ago

SkaematikBest ResponseYou've already chosen the best response.0
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 yintercept equal to logc a.
 one year ago

hbaBest ResponseYou've already chosen the best response.0
@Pauloading do you follow @Skaematik Method ?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
@Pauloading Do you want to follow my method ?
 one year ago

PauloadingBest ResponseYou've already chosen the best response.0
I'ts sound like greek to me :'( I'm really bad on it.
 one year ago

g152xxBest ResponseYou've already chosen the best response.0
if it is f1(a) when a = 2... set the first equation equal to 2, which will happen when x = 1. So, f1(2) = 1. Now if you want f1'(x), you have: = 1 / f'(f1(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.
 one year ago

hbaBest ResponseYou've already chosen the best response.0
@g152xx I hope you can help @Pauloading ,i gotta go workout ?
 one year ago

PauloadingBest ResponseYou've already chosen the best response.0
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)(xa) L(x) = f(a^3) + f'(3a^2)(a^31) L(x) = (1^3) + f'(3.1^2)(1^31) L(x) = 1 + 3.(0) L(x) = 1 Is it right?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
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.
 one year ago

g152xxBest ResponseYou've already chosen the best response.0
@Pauloading It sounds right 2me. Unless you wanted the answer in the form L(x) = f(a)+f'(a)(xa) instead of simplifying it.....
 one year ago

g152xxBest ResponseYou've already chosen the best response.0
@Pauloading Bump the question! Let's help him ppl! C'mon!
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
we have the formula, L(x) = f(a) + f'(a)(xa) why don't we use it directly ?? a=1 f(1) = 9 f'(1)=6 L(x) = 9+6(x1) = 6x +3
 one year ago

PauloadingBest ResponseYou've already chosen the best response.0
How did u get 6 at f'?
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
could you find f'(x) =... ?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
I asked her the same thing before :D
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
no, thats not correct... u need chain rule.
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
ok, what u got f'(x) =... ?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
The answer would be 2 (2 + 3 a^2) (4  2 a + a^3)
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
let me verify it for you....
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
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)(xa) L(x) = f(a^3) + f'(3a^2)(a^31) <INCORRECT L(x) = (1^3) + f'(3.1^2)(1^31) L(x) = 1 + 3.(0) L(x) = 1 L(x) = (1^3) + (3.1^2)(x1) =3+3(x1) =3x
 one year ago

PauloadingBest ResponseYou've already chosen the best response.0
Now I understood it '' Thanks everyone :D
 one year ago

hbaBest ResponseYou've already chosen the best response.0
d/da((a^32 a+4)^2) Using the chain rule, d/da((a^32 a+4)^2) = 2 u ( du)/( da), where u = a^32 a+4 and ( du^2)/( du) = 2 u: 2 (a^32 a+4) (d/da(a^32 a+4)) Differentiate the sum term by term and factor out constants: 2 (a^32 a+4) (d/da(a^3)2 (d/da(a))+d/da(4)) The derivative of 4 is zero: 2 (a^32 a+4) (d/da(a^3)2 (d/da(a))+0) The derivative of a is 1: 2 (a^32 a+4) (d/da(a^3)2 1) The derivative of a^3 is 3 a^2: Answer:   2 (a^32 a+4) (3 a^22)
 one year ago

hbaBest ResponseYou've already chosen the best response.0
I had to show her how to do it :)
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
ok,\[ \begin{array}l\color{red}{\text{P}}\color{orange}{\text{a}}\color{#e6e600}{\text{u}}\color{green}{\text{l}}\color{blue}{\text{o}}\color{purple}{\text{ }}\color{purple}{\text{C}}\color{red}{\text{e}}\color{orange}{\text{c}}\color{#e6e600}{\text{í}}\color{green}{\text{l}}\color{blue}{\text{i}}\color{purple}{\text{o}}\color{purple}{\text{}}\end{array} \]
 one year ago

hbaBest ResponseYou've already chosen the best response.0
\[\Huge{\bf{\color{blue}{O}\color{red}{K}}}\]
 one year ago
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