Question

I need help with linear approximation.

Answer 1

f(x)=the cube root of x at x=8. use the linear approximation to approximate the value of the cube root of 8.05 and the cube root of 25. I have no idea where they came up with the cube root of 8.05 or 25

Answer 2

They want you to use differentials.

Answer 3

\[
\Delta y \approx dy =y'dx=y'\Delta x
\]

Answer 4

IN this case :\[
\Delta x = 8.05 - 8
\]

Answer 5

I don't understand. Are you able to look at this example? I have not idea where they get cube root of 25 or the cube root of 8

Answer 6

http://tutorial.math.lamar.edu/Classes/CalcI/LinearApproximations.aspx

Answer 7

Well \(8=2^3\) so it is a perfect cube.

Answer 8

And since \(3^3=27\) they probably want you to use 27 to approximate 25

Answer 9

whats a differential?

Answer 10

Nevermind the differential. Let's just use tangent line approach.

Answer 11

\[
L(x) = f'(a)(x-a)+f(a)
\]

Answer 12

k

Answer 13

So \[
L(8.05)=f'(8)(8-8.05)+f(8)
\]