Question

Help with linearization problem but mostly teh differentiating part!! Attached is the problem

Answer 1

are u typing or is this s glitch?

Answer 2

\[L(x)\approx f(a)+f'(a)(x-a)\]
\[f(0)=\sqrt[5]{1+0}=1\]
\[f '(x)=\frac{ 1 }{5(1+x)^\frac{ 4 }{ 5 } }\]
\[f'(0)=\frac{ 1 }{ 5(1+0)^\frac{ 4 }{ 5 } }=\frac{ 1 }{ 5 }\]
0.95=1-.05 1.01=1+.01
x=-.05 x=.01

\[L(-.05) \approx 1+(\frac{ 1 }{ 5 })(-.05-0)=.99\]
\[L(.01)\approx 1+(\frac{ 1 }{ 5 })(.01-0)=1.002\]

Answer 3

So the equation F(x) equal to which equation? Becuase taht's what u use to approximate right?

Answer 4

srry i mean whats g(x)

Answer 5

g(x) would be f'(x) in what i showed above right?

Answer 6

oh ok i see!! THANKS FOR SUCH A THOROUGH EXPLANATION :D

Answer 7

You're welcome :) sorry i used f(x) instead of g(x).

Answer 8

If u dont mind can u plz explain how for the 2nd approximation of 1.1 u used 1 + 0.01. I thought it was supposed to be 1+ 0.1

Answer 9

you are right. its supposed to be 1+.1. So x=.1.

plug that into the equation and you should get the answer

Answer 10

ok i got 1.02

Answer 11

That's right. The exact value of \[\sqrt[5]{1.1}=1.019 or 1.02\]

Answer 12

thanks :)

Answer 13

You're welcome!

Answer 14

Wait srry but when i write the equation u gave for f'(X) into my hw it marks it wrong :(