Question

Use linear approximation to approximate √25.3 as follows.

Answer 1

The idea is to use the tangent line to $f(x)=\sqrt x$ at $x=25$ in order to approximate $f(25.3)=\sqrt{25.3}$.

We have

$f(x)=\sqrt x\implies f(25)=\sqrt{25}=5$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}$

so the linear approximation to $f(x)$ is

$L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92$

Hence $m=\frac1{10}$ and $b=\frac92$.

Then

$f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}$