EXAMPLE $1$ Find the linearization of the function $f(x)=\sqrt[\phantom{2}]{x+1}$ at $a=3$ and use it to approximate the numbers $\sqrt[\phantom{2}]{3\mathrm{.}95}$ and $\sqrt[\phantom{2}]{4\mathrm{.}05}.$ Are these approximations overestimates or underestimates?

SOLUTION The derivative of $f(x)=(x+1{)}^{\frac{1}{2}}$ is

$f^{\text{'}}(x)=$

and so we have $f(3)=2,$ and $f^{\text{'}}(3)=\frac{1}{4}.$ Putting these values into this equation, we see that the linearization is

$L(x)=f(3)+f^{\text{'}}(3)(x-3)=2$

$=2\mathrm{.}23,2\mathrm{.}26$

The corresponding linear approximation is

$\sqrt[\phantom{2}]{x+1}$~~$,$

In particular, we have

$\sqrt[\phantom{2}]{3\mathrm{.}95}$~~$\frac{5}{4}+\frac{{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}x}{4}=2\mathrm{.}23in(roundto$ four decimal places$)$

and

$\sqrt[\phantom{2}]{4\mathrm{.}05}$~~$\frac{5}{4}+\frac{1}{4}$ Your answer $is$ incorrect.

The linear approximation is illustrated in the figure to the left. We see that, indeed, the tangent line approximation is a good approximation to the given function when $x$ is near $3.$ We also see that our approximations are overestimates because the tangent line lies above the curve.

Of course, a calculator could give us approximations for $\sqrt[\phantom{2}]{3\mathrm{.}95}$ and $\sqrt[\phantom{2}]{4\mathrm{.}05},$ but the linear approximation gives an approximation over an entire interval. Help me fill in the blanks that are wrong