# How do you use linear approximation to the square root function to estimate square roots sqrt 8.95?

Dec 3, 2016

The linear approximation of $f$ at $a$ is $L \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$. This is one way of writing the equation of the line tangent to the graph of $f$ at $\left(a , f \left(a\right)\right)$

#### Explanation:

The function we want to approximate is $f \left(x\right) = \sqrt{x}$.

We want to estimate $f \left(8.95\right) = \sqrt{8.95}$ so we need a value for $a$ that is close to $8.95$ and for which we can find $f \left(a\right)$. We can readily find $f \left(1\right)$ or $f \left(4\right)$, but $f \left(5\right)$ would be more challenging.

Although we can easily find $f \left(25\right)$, we'd like $a$ to be as close as we can get to 8.95 and still calculate $f \left(a\right)$.

By now you may be anxious for me to say, "we'll use $a = 9$". So I will. We'll use $a = 9$.

So $f \left(a\right) = f \left(9\right) = 3$

For the linearization we'll need $f ' \left(9\right)$ as well.

$f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$, so $f ' \left(9\right) = \frac{1}{6}$

The linearization of $f \left(x\right) = \sqrt{x}$ at $a = 9$ is

$L \left(x\right) = 3 + \frac{1}{6} \left(x - 9\right)$

Our estimate is

$f \left(8.95\right) \approx L \left(8.95\right) = 3 + \frac{1}{6} \left(8.95 - 9\right)$

$= 3 - \frac{0.05}{6} = 3 - 0.008 \overline{3}$

$= 2.991 \overline{6}$

Round to the desired accuracy. Keep in mind that this is an estimate for the desired number $\sqrt{8.95}$ so don't claim too much accuracy.

$2.99$ or perhaps even $2.991$ seems reasonable. (Although $2.9917$ is accurate rounded to 4 decimal places.)