Find the linearization of the function $f(x)=\sqrt[\phantom{2}]{x1}$ at $a=8$ and use it to approximate the numbers $\sqrt[\phantom{2}]{8\mathrm{.}95}$ and $\sqrt[\phantom{2}]{9\mathrm{.}04}.$ Are these approximations overestimates or underestimates?

Solution

The derivative of $f(x)=(x1{)}^{\frac{1}{2}}$ is

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

and so we have $f(8)=$ and $f^{\text{'}}(8)=\frac{1}{6}q,.$ Putting these values into the equation $L(x)=f(a)f^{\text{'}}(a)(x-a),$ we see that the linearization is

$L(x)=f(8)f^{\text{'}}(8)(x-8)=()(x-8)$

$$

The corresponding linear approximation is

$\sqrt[\phantom{2}]{x1}$~~ $\frac{x}{6},($ when $xis$ near $8)$

In particular, we have

$\sqrt[\phantom{2}]{8\mathrm{.}95}$~~$\frac{5}{3}\frac{0\mathrm{.}95,x}{6}=1\mathrm{.}8250,x,(roundto$ four decimal places$)$

and

$\sqrt[\phantom{2}]{9\mathrm{.}04}$~~$\frac{5}{3}\frac{1\mathrm{.}04,x}{6}=1\mathrm{.}8400,x,(roundto$ four decimal places$)$

The linear approximation is illustrated in the figure below.

$$

The corresponding linear approximation is

$\sqrt[\phantom{2}]{x1}$~~ $\frac{x}{6},($when $x$ is near $8).$

In particular, we have

$\sqrt[\phantom{2}]{8\mathrm{.}95}$~~$\frac{5}{3}\frac{0\mathrm{.}95,x}{6}=1\mathrm{.}8250,x(roundto$ four decimal places$)$

and

$\sqrt[\phantom{2}]{9\mathrm{.}04}$~~$\frac{5}{3}\frac{1\mathrm{.}04,x}{6}=1\mathrm{.}8400,x(roundto$ four decimal places$).$

The linear approximation is illustrated in the figure below.

We see that, indeed, the tangent line approximation is a good approximation to the given function when $x$ is near $8.$ We also see that our approximations are overestimates because the tangent line lles above the curve.

Of course, a calculator could give us approximations for $\sqrt[\phantom{2}]{8\mathrm{.}95}$ and $\sqrt[\phantom{2}]{9\mathrm{.}04},$ but the linear approximation gives an approximation over an entire interval. Can I get help on why my answers are wrong and the exact steps. I am confused as to why those boxes were marked as incorrect.