(Solved): EXAMPLE 1 Find the linearization of the function f(x)=x+1 at a=3 and use it to approximate the ...

EXAMPLE 1 Find the linearization of the function $f(x)=\sqrt{x+1}$ at $a=3$ and use it to approximate the numbers $\sqrt{3.99}$ and $\sqrt{4.04}$. Are these approximations overestimates or underestimates? SOLUTION The derivative of $f(x)=(x+1{)}^{1/2}$ is $f^{?}(x)=\sqrt{\frac{1}{2\sqrt{x+1}}}$ and so we have $f(3)=$ and $f^{?}(3)=$. Putting these values into this equation, we see that the linearization is $L(x)=f(3)+f^{?}(3)(x?3)=?+(x?3)$ $=$ The corresponding linear approximation is $\sqrt{x+1}?+\frac{x}{4}(\text{ when }x\text{ is near }3)$ In particular, we have $\begin{array}{c}\sqrt{3.99}?\frac{5}{4}+\frac{}{4}=\text{ (round to four decimal places) }\\ \text{ and }\\ \sqrt{4.04}?\frac{5}{4}+{!}_4=\text{ (round to four decimal places). }\end{array}$ 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 $\sqrt{3.99}$ and $\sqrt{4.04}$, but the linear approximation gives an approximation over an entire interval.