Question

Use linear approximation, i.e. the tangent line, to approximate √ 16.2 as follows: Let f ( x ) = √ x . Find the equation of the tangent line to f ( x ) at x = 16 L ( x ) = Using this, we find our approximation for √ 16.2 is NOTE: For this part, give your answer to at least 9 significant figures or use an expression to give the exact answer. Get help: Video

Answer 1

The linear approximation to $f(x)$ near $x=16$ is

$L(x)=f(16)+f'(16)(x-16)$

We have

$f(x)=\sqrt x\implies f(16)=4$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(16)=\dfrac18$

$\implies L(x)=4+\dfrac18(x-16)=\dfrac x8+2$

Then

$f(16.2)\approx L(16.2)\iff\sqrt{16.2}\approx\dfrac{16.2}8+2$

$\implies\boxed{\sqrt{16.2}\approx4.025=\dfrac{161}{40}}$