estimate $f(3.8)-f(3)$ with $f(x) = \sqrt{x+1}$

Question

We are given $f(x)= \sqrt{x+1}$ and use the Linear Approximation to this function at $a=3$ with $\triangle x = 0.8$ to estimate $f(3.8)-f(3) = \triangle f \approx df$

Can you guys explain each step of the process because I am having a hard time understanding how to use the Linear Approximation equation.

Answer 1

It's really just an application of linear approximation using Taylor's theorem $$ f(3.8)\approx f(3)+.8f'(3)\implies f(3.8)-f(3)\approx .8f'(3) $$ Can you finish the problem? The remaining steps involve computing a derivative and plugging numbers into the argument of your function.

Answer 2

By the Lagrange formula (https://en.wikipedia.org/wiki/Mean_value_theorem) we obtain: $$\sqrt{1+3.8}-\sqrt{1+3}=\frac{0.8}{2\sqrt{1+(3+0.8\theta)}}<0.2,$$ where $\theta\in(0,1)$.