If the function of $f:[0,4]\to R$ is differentiable, then $\left(f\right(4){)}^2-(f\left(0\right){)}^2=?$

Given function $f:\left[0,4\right]\to R$ is differentiable in given interval.

 Here, $f$ is differentiable $\Rightarrow f$ is also continuous.  

Lagrange's mean value theorem: Let a function f be defined such that $f:[a,b]\to R$ be a continuous function on $[a,b]$ and differentiable on $(a,b)$. Then there exists some a point c in this interval$(a,b)$ such that the derivative of the function at the point $c$ is equal to the difference of the function values at these points, divided by the difference of the point values.

$\Rightarrow f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Now, by Lagrange's mean value theorem, there exist $a\in (0,4)$ such that

$f^{\text{'}}\left(a\right)=\frac{f\left(4\right)-f\left(0\right)}{4-0}=\frac{f\left(4\right)-f\left(0\right)}{4}$

Intermediate Mean value theorem: If a function$f\left(x\right)$ that is continuous on an interval $\left[a,b\right]$, then for every $y$-value between $f\left(a\right)$ and $f\left(b\right)$, there exists some $x$-value in the interval $\left(a,b\right)$.

Also, by intermediate mean value theorem, there exists $b\in (0,4)$ such that

$f\left(b\right)=\frac{f\left(4\right)+f\left(0\right)}{2}$

By above both equations we get,

$$\begin{gathered} \Rightarrow f^{\text{'}}\left(a\right)f\left(b\right)=\frac{f\left(4\right)-f\left(0\right)}{4}\times \frac{f\left(4\right)+f\left(0\right)}{2} \\ \Rightarrow f^{\text{'}}\left(a\right)f\left(b\right)=\frac{\left(f\right(4){)}^2-(f\left(0\right){)}^2}{8} \\ \Rightarrow \left(f\right(4){)}^2-(f\left(0\right){)}^2=8f^{\text{'}}\left(a\right)f\left(b\right) \end{gathered}$$

Therefore, the correct answer is option $2.$