Linear Approximations — Question 10

Question 10 - Solution

For $f(x)=\sqrt{x}$, use the nearby perfect square $a=9$.

$$f(9)=3,\qquad f'(9)=\frac16,\qquad L(x)=3+\frac{x-9}{6}.$$

Therefore

$$\boxed{\sqrt{10.2}\approx L(10.2)=3.2.}$$

Since $f''(x)=-1/(4x^{3/2})<0$ for $x>0$, the tangent line lies above the curve, so this is an overestimate.

For comparison, $\sqrt{10.2}\approx3.193743885$; the error is approximately $0.006256115$.