Functions and More Functions

$f(\sqrt{6}) = \frac{3}{5} (\sqrt{6}) +\sqrt{3}$

$f(\sqrt{8})= \frac{3}{5}(\sqrt{8})+\sqrt{3}$

$f(\sqrt{6})-f(\sqrt{8})=\frac{3}{5}(\sqrt{6})+\sqrt{3}-\frac{3}{5}(\sqrt{8})-\sqrt{3}\\=\frac{3}{5}(\sqrt{6})-\frac{3}{5}(\sqrt{8})\\=\frac{3}{5}(\sqrt{6}-\sqrt{8})$

Substitute this for $f(\sqrt{6})-f(\sqrt{8})$ in $\frac{f(\sqrt{6})-f\sqrt{8})}{\sqrt{6}-\sqrt{8}}$ :

$\frac{\frac{3}{5}(\sqrt{6}-\sqrt{8})}{\sqrt{6}-\sqrt{8}}\\=\frac{3}{5}$

The required expression reminds us of the notion of the $\textit {Derivative}$ at any point $x$ which is: $\frac{d(f(x))}{dx}=\lim_{y\to x} \frac{f(y)-f(x)}{y-x}$

But here in our required expression the limit part is missing. The essence of this limit part is that $y$ is very much close to $x$ because the derivative at $x$ is the slope of the $\textbf {tangent}$ drawn at $(x,f(x))$ .

But here since $f(x)$ is a $\textbf {linear }$ function so the derivative is constant that means $\textbf {there is no need of the limit part here.}$

Hence $\frac{f(√6)-f(√8)}{√6-√8}=f'(x)=\boxed{\frac{3}{5}}$

Given situation can be taken as two coordinate points given on the curve: ( sqrt6, f(sqrt6)) and (sqrt8, f(sqrt8)). Consider these set of coordinates as (x1, y1) and (x2, y2) Then given expression is nothing but the slope of the chord joining these two points. that can be evaluated directly.

This can also be done through mean value theoreum.

I used probability with instincts in my guesses... hahaha... Who would have thought that guessing answers to difficult problems would be fun....