Question

Find the linerization of a suitable function and then use it to aproximate the number:
(i)1.002^3 (ii)63.8^(1/2) (iii) 31.08^(1/5) (iv) sin 0.1
teach showed me an example but it wasn't like this

Answer 1

Linearization is
L(x)=y(x)+y'(x)dx

Answer 2

i)1.002^3 Something cube
Let say x^3
Dy/dx (x^3) =3x^2

Answer 3

L(x)=y(x)+y'(x)dx
y(x)=x^3
y'(x)=3x^2

Answer 4

Let's plug them in
x^3+3x^2 dx

Answer 5

any question so far?

Answer 6

yea
that use a suitable function part, so all thats asking is use that L(x) formula?

Answer 7

i ask because in the example she game us a f(x) formula with it

Answer 8

and a point of x

Answer 9

So " use a suitable function part"
Some example
2^3 -- use this function x^3
Ln(3) -- use this func ln(x)

Answer 10

y=(32-x)^(1/5)
y'=-1/5(32-x)^(-4/5)
y'=-1/5(32-0)^(-4/5)=-1/5(32)^(-4/5)=-1/5(1/16)=-1/80
y=mx+b
y=-1/80*x+b (we know a point on the lint (0,2)
y=-1/80*x+2
L(x)=mx+b=-1/80*x+2
so we want to find L at x=.92=92/100=23/25
L(23/25)=-1/80*23/25+2=3977/2000=1.9885
if we have
done y=(32-.92)^(1/5)=(31.08)^(1/5)=1.988365 approximately
all you are doing is finding the tangent line

Answer 11

wow, lol gimmie a min to try and digest this, u guys are good,

Answer 12

wow, didn't even figure all that formula was just for the tangent line, with all the f' of tyhis and that

Answer 13

thnx alot for the help, its much appreciated guys