Question

This is a tough question for Honors Calculus. Someone please help

Answer 1

Let
`f(x)=x^(1/4)=\sqrt[4]x`. We have
`f(16)=2`, since
`2^4=16`. We can approximate values of
`f(x)` for
`x` near
`16` with the linear approximation granted by the mean value theorem: For some
`x` near
`a`,

`f'(a)\approx(f(x)-f(a))/(x-a)\implies f(x)\approx f(a)+f'(a)(x-a)`

We have

`f(x)=x^(1/4)\implies f'(x)=\frac1{4x^(3/4)}\implies f'(16)=\frac1{32}`

Then

`16.5^(1/4)\approx16^(1/4)+f'(16)(16.5-16)\implies16.5^(1/4)-16^(1/4)\approx(0.5)/(32)=\frac1{64}=0.015625`

That is, the approximate difference between
`16.5^(1/4)` and
`16^(1/4)=2` is 0.015625.

Checking with a calculator, you'll find the difference to be about 0.0154452.