Question

given that y = x^(1/3), use calculus to determine an approximate value for 8030^(1/3).

pls show working.

Answer 1

why would you need calculus?

cubed root of 8030 is approx 20.025

Answer 2

Using linear approximation
\[L(a)=f(a)+f'(a)(x-a)\]
Taking a to be 8000
(because we know the cube root of 8000 and it's close to 8030)
\[y=f(x)=x^{1/3}\]
\[f'(x)=1/(3x^{2/3})\]
\[L(8000)=f(8000)+f'(8000)(x-8000)\]
\[L(8000)=(8000)^{1/3}+1/(3(8000)^{2/3})(x-8000)\]
\[L(8000)=20+(1/1200)(x-8000)\]
\[L(8000)=20 +x/1200-20/3\]
Thus the linear approximation of y=x^1/3 near 8000 is
\[L(x)=40/3+x/1200\]
\[L(8030)=40/3+8030/1200\]
Which is 20.025

Answer 3

thats would be 2.0

Answer 4

you dumb cow, the point of the question is to show we know calculus. not to prove that we know how to use a calculator.