Question

Hello, please help me with this question
Linear derivation. Photo attached

Answer 1

\[f(x)=x^3\\f(2)=2^3=8\]
we need to find \(f(2.3)=f(2+0.3)\)
so, for a small change in x, what is the change in f?

Answer 2

from\[f'(x)=3x^2\\\Delta f(x)=3x^2\Delta x\\
\text{small change near x=2}\\
\Delta f(2)=3(2)^2\Delta x=12\Delta x\]
now, \(\Delta x=0.3\)
\[\Delta f(2)=12(0.3)=3.6\]

Answer 3

now, \[f(2.3)=f(2+0.3)\approx f(2)+\Delta f(2)=8+3.6=\boxed{11.6}\]

Answer 4

they want you to use \(x=3\) and \(\Delta x=-0.4\)

Answer 5

why do you use 2 from the first step?

Answer 6

because that is the center of the region of our approximation.
we are approcimating near x=2

Answer 7

Oh Isee. not 2.6?

Answer 8

you can approximate from x=3 too

Answer 9

I see. M is not 1.6 isnt?

Answer 10

the idea is to have dx to be as small as possible.

Answer 11

Ok...

Answer 12

what shall i do in order to solve ths probelm?

Answer 13

same steps as I did but use x="3" and dx=-0.4 (because 3-0.4=2.6)

Answer 14

Im lost :/

Answer 15

\[
f(2.6)=f(3+(-0.4))=f(3)+f'(3)\Delta x\\
f(2.6)=(3)^3+3(3)^2(-0.4)\\
f(2.6)=27-3.6\\
f(2.6)=13.4
\]

Answer 16

ok is this for solpe? m or b?

Answer 17

slope is f'(3)

Answer 18

mmm 3 is the answer ? why is this...

Answer 19

?

Answer 20

3 is m?

Answer 21

no
\[f'(x)=3x^2\\
m=f'(3)=27\]

Answer 22

Ooo.. I got it.
what about b?

Answer 23

now,we have\[y=mx+b\\
y=27x+b\]
we know that when x=3, y=27
plug it in
\[27=27(3)+b\]
solve for b

Answer 24

54?

Answer 25

noo. try again.. solve for "b"

Answer 26

27=81+b
27-81=B
-54=b?

Answer 27

ok.. b= -54
so, now we have \[y=27x-54\]

Answer 28

I got it!

Answer 29

to estimate f(2.6),
put x=2.6 and solve for "y"

Answer 30

27(2.6)-54?

Answer 31

yep

Answer 32

16.2?

Answer 33

yep

Answer 34

Thank you so much for clear explanation!!

Answer 35

could you help me other question? i will open up new wndow

Answer 36

yer welcome.