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 $a$,

$f'(a)\approx\dfrac{f(x)-f(a)}{x-a}\implies f(x)\approx f(a)+f'(a)(x-a)$

We have

$f(x)=x^{1/4}\implies f'(x)=\dfrac1{4x^{3/4}}\implies f'(16)=\dfrac1{32}$

Then

$16.5^{1/4}\approx16^{1/4}+f'(16)(16.5-16)\implies16.5^{1/4}-16^{1/4}\approx\dfrac{0.5}{32}=\dfrac1{64}=0.015625$

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.