Question

Calculus help!

Answer 1

Find the linear approximation of the function g(x) = cuberoot(1+x) at a = 0 and use it to approximate the numbers cuberoot(0.9) and cuberoot(1.1). Illustrate by graphing g and the tangent line.

Answer 2

So, here is what I have so far:
g(0) = 1
g'(x) = 1/3(1+x)
g'(0) = 1/3
L(x) = 1 + 1/3x

Answer 3

Your deriv isn't correct. Where's the exponent?

Answer 4

Now, to find the approximate value of cuberoot(0.95) would I plug in 1.95 for x in the linearization equation?

**it is supposed to be cuberoot(0.95) not (0.9)**

Answer 5

Crap! Hold on. Good catch agent.

Answer 6

Agent, I did a problem very similar to this recently but I'm not sure if it's right... mind popping over to check it out? I did the same process but forgot the exponent on the derivative. No one caught it :O
http://openstudy.com/study#/updates/57d738c9e4b0067db6b6350c

Answer 7

Looks about right except your deriv was wrong like you said.

Use L(x) = f(a) +f'(a) *(x-a) which is really a fancy form of y - y1 = m(x - x1)

Answer 8

\[\large g(x) =\sqrt[3]{1+x}\]\[\large g'(x) =\frac{ 1 }{ 3(1+x)^{2/3} }\]g(0) = 1
g'(0) = 1/3

L(x) = 1 + 1/3x

So you were correct.

Answer 9

Awesome, I think I've got it! Thanks Angel :D

Answer 10

Welcome pebbles!

Answer 11

And you had it already, without my help :D