How do you use local linear approximation to approximate the value of the given quantity to $4$ decimal places ${(80.5)^{\dfrac{1}{4}}}$ ?

Hint: Local approximation is a method of approximation that relies on information about the function's value and derivatives at the same point. The goal is to get a close approximation of the function in the vicinity of the benchmark point.

Complete step by step answer:
Let $f(x) = {x^{\dfrac{1}{4}}}$ and we’ll use it in Taylor series with $x = a$ considering only the first term, so as to have a function with a linear approximation.
Consider,
$f(a) = {a^{\dfrac{1}{4}}}$
On differentiating, we get,
$f'(x) = \dfrac{1}{4}{x^{ - \dfrac{3}{4}}}$ so $f'(a) = \dfrac{1}{4}{a^{ - \dfrac{3}{4}}}$
We know that,
${x^{\dfrac{1}{4}}} = \sqrt[4]{a} + \dfrac{{x - a}}{{4\sqrt[4]{{{a^3}}}}} + {R_2}(x)$
Where, ${R_2}(x)$ is the sum total of the terms we ignore and it is known as the approximation error.
Now, on considering $a = 81$, we get,
$\sqrt[4]{{81}} = 3$ and $\sqrt[4]{{{{81}^3}}} = 27$
Therefore,
${(80.5)^{\dfrac{1}{4}}} \cong \dfrac{{80.5 - 81}}{{4 \times 27}} = 3 - \dfrac{{0.5}}{{108}} \cong 2.9953$
To check whether the approximation is correct to $4$ decimal places, we use Lagrange’s formula for maximum error:
$\left| {{R_2}(x)} \right| \leqslant \dfrac{{{{(x - a)}^2}}}{2}\mathop {\max }\limits_{\xi \in (x,a)} \left| {f''(\xi )} \right|$
Now, we evaluate the ${2^{nd}}$ derivative,
$f''(x) = - \dfrac{3}{{16}}{x^{ - \dfrac{7}{4}}}$
We can notice that $\left| {f''(x)} \right|$ is a strictly decreasing function. Hence, in the interval $x \in (80.5,81)$ , the maximum value obtained for $x = 80.5$ .
We can write ${(80.5)^{ - \dfrac{7}{4}}}$ as,
${(80.5)^{ - \dfrac{7}{4}}} = \dfrac{1}{{{{\left( {{{(80.5)}^{\dfrac{1}{4}}}} \right)}^7}}}$
Using the value of approximation which we already found, we can write it as,
${(80.5)^{ - \dfrac{7}{4}}} \cong \dfrac{1}{{{{\left( {2.9953} \right)}^7}}}$
This signifies that for a good approximation the approximation error is lesser than:
$\left| {{R_2}(x)} \right| \leqslant \dfrac{{{{(80.5 - 81)}^2}}}{2}\left( {\dfrac{3}{{16}}} \right) \times \dfrac{1}{{{{(2.9953)}^7}}} = 0.00000541$
This provides us a good assurance that our approximation is good to $4$ decimal places.

Note: Linear approximation, also known as linearization, is a technique for estimating the value of a function at a given point. Linear approximation is useful since finding the value of a function at a particular stage can be difficult.
A good example of this is square roots.