Question

Find the linearization function and use this function to approximate cuberoot(126). show work to ear credit. Round the approximation to 2 decimal places.

having trouble finding L(x) and roughly what cuberoot(126) is = to

Answer 1

First: the required equation for the "linearization" of a function f(x) is\[f(x)~\approx~equal~\to~f(a)+f '(a)*(x-a)\]

Answer 2

In other words, at the x value 'a,' the function f(x) is approximately equal to the value of f(x) at 'a' PLUS the product of the derivative of f(x), evaluated at 'a', times (x-a).

Here we're asked to approx. the cube root of 126.

My first question for you is this: What is the integer cube closest to 126? Note that 120, 121, ...124 are not perfect cubes.

Answer 3

heres what i have so far but i'm going wrong somewhere f(x) = cuberoot(x) = x^1/3 then f prime of x i get 1/3x^-2/3 = 1/3cuberoot(x^2) so then with all this L(x) = f(5) +f^i(5)(x-5) its simplifying that..thats my issue

Answer 4

Thanks for sharing your work!

I'd also choose f(x) to be the cube root function:

f(x) = x^(1/3). Then the derivative, f(x), is \[f '(x)=\frac{ 1 }{ 3 }x ^{-2/3}\]

Answer 5

Could you consider why you used a=5? What was your justification? Note that the closest perfect cube is 125. I'm using a=125, not a=5.

Answer 6

Your approximation is then (before substituting a=125):\[f(x)\approx f(a)+f '(a)(x-a).\]Now, please substitute a=125 and simplify the result.

Answer 7

Both f(a) and f '(a) are easily evaluated. We want to approx. the cube root of 126. Substituting 126 for x, and subtracting a = 125 from that, we get

Answer 8

\[f(126) \approx f(125)+f'(125)(126-125)\]

Answer 9

Please finish this. Check your result by finding the cube root of 126 on a calculator.