Use a linear approximation (or differentials) to estimate the given number. 99.8

Question

Novembruuo · Accepted Answer

Approximation can be done by using a linearized model of our function:

L (x) \approx f (a) + f^{'} (a) \cdot (x - a)

Now we need to find the derivative of f(x):

f (x) = \sqrt{x}

f^{'} (x) = {(x^{\frac{1}{2}})}^{'} = \frac{1}{2} x^{(\frac{1}{2} - 1)} = \frac{1}{2} x^{- \frac{1}{2}}

= \frac{1}{2 \sqrt{x}}

Without a calculator we don't know the value of

\sqrt{99.8}

, but since we need an approximation, we'll use the closest value that we know and that's

\sqrt{100}

. Hence:

f (x) = \sqrt{x} = \sqrt{100} = 10

f^{'} (x) = \frac{1}{2 \sqrt{x}} = \frac{1}{2 \sqrt{100}} = \frac{1}{2 \cdot 10}

= \frac{1}{20}

Now we can calculate the approximation using the linearized model:

L (x) \approx f (a) + f^{'} (a) \cdot (x - a)

L (99.8) = \sqrt{99.8} \approx f (100) + f^{'} (100) \cdot (99.8 - 10)

\approx 10 + \frac{1}{20} \cdot (44 - 45)

\sqrt{99.8} \approx 9.95

Result:

\sqrt{99.8} \approx 9.95

Use a linear approximation (or differentials) to estimate the given