How do you use linear approximation about x=100 to estimate #1/sqrt(99.8)#?

Question

Cameron Alexander · Accepted Answer

#f(x) = 1/sqrtx#
#f(x) = x^(-1/2)# #f(100) = 1/sqrt100 = 1/10 = .1# #f'(x) = (-1/2)(x^(-3/2))#
#f'(100) = (-1/2)(100)^(-3/2) = -.0005# #L(x) = -.0005(x-100)+.1#
#L(99.8) = -.0005(-.2)+.1 = .1001#

Elijah Abram · Answer

undefined To use linear approximation about $x = 100$ to estimate $1/\sqrt{99.8}$, you first find the linear approximation of the function $f(x) = 1/\sqrt{x}$ near $x = 100$. The linear approximation is given by the formula:

$$L(x) = f(a) + f'(a)(x - a)$$

Where $a$ is the point of approximation, $f(a)$ is the value of the function at $x = a$, and $f'(a)$ is the derivative of the function evaluated at $x = a$.

First, find the value of $f(100)$ and $f'(100)$:

$$f(100) = \frac{1}{\sqrt{100}} = \frac{1}{10}$$
$$f'(x) = -\frac{1}{2x^{3/2}}$$
$$f'(100) = -\frac{1}{2(100)^{3/2}} = -\frac{1}{200}$$

Now, plug these values into the linear approximation formula:

$$L(x) = \frac{1}{10} - \frac{1}{200}(x - 100)$$

Finally, plug in $x = 99.8$ to get the estimate:

$$L(99.8) = \frac{1}{10} - \frac{1}{200}(99.8 - 100)$$
$$L(99.8) = \frac{1}{10} - \frac{1}{200}(-0.2)$$
$$L(99.8) = \frac{1}{10} + \frac{1}{1000}$$
$$L(99.8) = \frac{100 + 1}{1000}$$
$$L(99.8) = \frac{101}{1000}$$

So, the estimate of $1/\sqrt{99.8}$ using linear approximation about $x = 100$ is $101/1000$.