Question

Linear approximation or differentials. :P

Answer 1

\[\sqrt[3]{1001}\]

Answer 2

\[f(x)=\sqrt[3]{x},,, a=1000\]

Answer 3

f(a)+f '(a)(x-a)

Answer 4

\[10+\frac{ 1 }{ 300 }(\sqrt[3]{1001}-1000)\]

Answer 5

=6.700

Answer 6

the answer is 10.003 where am i going wrong?

Answer 7

hello

Answer 8

\[f(x)=x^{1/3}\]\[f(x)\approx f(x_0)+f'(x_0)(x-x_0)\]let\[x_0=1000\]\[f(x)\approx10+\frac13(1000)^{-2/3}(x-1000)=10+\frac x{300}-\frac{10}3\]so about \(x_0=1000\) we have\[f(x)\approx\frac{20}3+\frac x{100}\]

Answer 9

I just brushed up on this stuff so I thought I'd give it a try :)

Answer 10

ahhh, ok so at the end plug in 1000

Answer 11

no, plug in x=1001

Answer 12

you need to choose \(x\approx x_0=1000\) and \(1001\approx1000\), so it works

Answer 13

that does not equal 10.003 though :(

Answer 14

it does by my calculator :/

Answer 15

oh I made a typo at the end

Answer 16

\[\frac{ 20 }{ 3 } + \frac{ 1001 }{ 100 }\]

Answer 17

I should have gotten\[f(x)\approx\frac{20}3+\frac x{300}\]I switched out the base on x for some reason, you should really check my work though

Answer 18

ahh