Linear Approximations — Question 8

Question 8

Use a linear approximation to estimate $\sqrt{10.2}$.

• Define an appropriate function $f(x)$ and choose a point $a$ close to 10.2 where $f(a)$ is easy to compute.

• Find the linear approximation $L(x)$ to $f(x)$ at $x = a$.

• Use the approximation to estimate $\sqrt{10.2}$, and compare it with the actual value.

Question 8 - 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 $0.006256115$.